IB - Chapter 8 Coordinate Geometry Part 1

Page 1 of 42

Chapter 8 Coordinate Geometry

Part 1

8A Equation of Lines

8B DMS Formulas Intro – Find Midpoint

8C DMS – Find Endpoint

8D DMS – Apply Formula

8E Perpendicular Bisector

8F Irregular Shapes: Use Pythagorean Theorem

8G Partition a Line

•Two lines are parallel lines if they do not intersect. The slopes of the lines are the same.

•f(x)=m1x+b1 and g(x)=m2x+b2 are parallel if m1=m2

•If and only if b1=b2 and m1=m2, we say the lines coincide. Coincident lines are the same line.

•Two lines are perpendicular lines if they intersect at right angles.

•f(x)=m1x+b1 and g(x)=m2x+b2 are perpendicular if m1∗m2=−1, and m2= (-1)/m1

•HOW TO: GIVEN THE EQUATION OF A LINEAR FUNCTION, WRITE THE EQUATION OF A LINE WHICH PASSES THROUGH A GIVEN POINT AND IS PARALLEL TO THE GIVEN LINE.

•Find the slope of the function.

•Substitute the slope and given point into point-slope or slope-intercept form.

•Simplify.

•A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not –1. Doesn’t this fact contradict the definition of perpendicular lines?

•No. For two perpendicular linear functions, the product of their slopes is –1. However, a vertical line is not a function so the definition is not contradicted.

•HOW TO: GIVEN THE EQUATION OF A LINEAR FUNCTION, WRITE THE EQUATION OF A LINE WHICH PASSES THROUGH A GIVEN POINT AND IS PERPENDICULAR TO THE GIVEN LINE.

•Find the slope of the given function.

•Determine the negative reciprocal of the slope.

•Substitute the new slope and the values for x and y from given point into g(x)=mx+b.

•Solve for b.

•Write the equation of the line.

•HOW TO: GIVEN TWO POINTS ON A LINE AND A THIRD POINT, WRITE THE EQUATION OF THE PERPENDICULAR LINE THAT PASSES THROUGH THE POINT.

•Determine the slope of the line passing through the points.

•Find the negative reciprocal of the slope.

•Use slope-intercept form or point-slope form to write the equation by substituting the known values. & Simplify.

•

Page 2 of 42

8A Equ of Parallel and Perpendicular Lines

EQUATIONS OF LINES Write an equation for the following lines:

•1) For the line that passes through (4,-8) and has a slope of -2.

•Answer :

In slope intercept form

The equation of a line is y=mx+c where m is slope and c is y-intercept.

slope m = 2 then partial equation is y = -2x + c

The line passes through ( 4,-8)

y = -2x + c

-8 = -8 + c

c = 0

Thus equation is : y = -2x

Another method

y-y1 = m (x-x1 )

y+8=-2(x-4)

y+8=-2x + 8

y=-2x.

•

X	0	3
Y	0	-6

2) For the line that has a slope 2/5 of passes through the point (-10,4).

•Answer:

•slope m = 2/5 then partial equation is y = 2/5 x + c

The line passes through ( -10,4)

•

y = 2/5 x + c

•

4 = 2/5 (-10) + c

•c = 8

Thus equation is : y = 2/5 x +8

•Another method

•

y-y1 = m (x-x1 )

y - 4= 2/5 (x + 10)

•y - 4 = 2/5 x + 4

•y= 2/5 x + 8

•

X	0	5
Y	8	10

3. For the line that has zero slope and passes through the point (1, -2)

Answer:

slope m = 0 then partial equation is y = 0 x + c

The line passes through ( 1,-2)
y = 0x + c
-2 = 0 (1) + c

c = -2
Thus equation is : y = -2

Another method

y-y1 = m (x-x1 )
y + 2= 0 (x -1)

y + 2 = 0

y = -2

4. For the line that has undefined slope and passes through the point (-3,-7).

If the slope of the line is undefined, then, by definition the line is a vertical line. For a vertical line, the value of x is the same for each and every value of y. A Vertical line with an undefined slope does not have a y intercept. So a line with an undefined slope has the equation of the type x = a where a is the x intercept.

•If line is vertical then it is parallel with y-axis & true for every constant x = - 3

•Slope = Rise/Run is undefined

• if Run = 0. Since you can’t divide by 0. This is, therefore, a vertical line.

•The vertical line goes through (-3,-7)

•Therefore: the equation is: x = -3

•The y-coordinate can be any Real Number but x-coordinate can only be -3 making the Run = 0 and the slope undefined.

•

•Another method

•y-y1 = m (x-x1 )

y + 7= m (x +3)

•m = (b+7)/(a+3)

•y + 7 * (a+3)/(b+7) = (x +3)

•As a+3=0 and 0 divided by anything is 0, so our equation is,x+3=0 i.e. x= -3

• x = -3

5. For the line parallel to the line 4x-6y=3 and whose y-intercept is 9.

•Answer:

•

The line 4x - 6y =3

•6y = 4x - 3

y = 4/6 x - 3/6

y = 2/3 x - 1/2

•c = 9

•y = mx + c

• y = 2/3 x + 9

•

X	0	-3
Y	9	7

6. For the line parallel to the line y-5x=4 passes through the point (1,3).

•Answer:

•The line y – 5x = 4

•y = 5x + 4

•m = 5

y-y1 = m (x-x1 )

y - 3= 5 (x - 1)

•y - 3 = 5 x - 5

•y= 5x -5 +3

•y = 5x -2

X	0	1	2
Y	-2	3	8

7. For the line parallel to the line x = 12 and and passes through the point (-3,4)

•Answer:

•The slope of line x=12 is infinite

•The equation of line passing through point (-3,4) and having slope infinite is y−4=1/0(x+3)Which implies x+3=0 , which gives x=-3

•x=12 is a line parallel to the y-axis passing through all points in the plane with an x-coordinate of 12

•therefore a line parallel to x=12 will also be a line parallel to the y-axis and passing through all points in the plane with the same x-coordinate

•the line required passes through the point (−3,4)

•⇒equation of parallel line is x=−3

X	-3	12
Y	0	0

8. For the line parallel to the line y= -6 and passes through the point (4,-5)

•Answer:

•y = -6 is a straight line with a slope of zero.

•Any line of the form y = n will have the same slope. the one passing through (4,-5) is: y = -5

•

•y = mx+c

•-5 = 0(4) + c

•-5 = 0 + c

•c = -5

•y = 0x - 5

•y = -5

•

•y - (-5) = 0(x - (4))

•y + 5 = 0(x + 4)

•y + 5 = 0

•y = -5

X	0	0
Y	-5	-6

Page 4 of 42

9. What is the equation of a line that is parallel to the line y = - 2x + 8 and passes through the point (3,6)?.

•Answer:

•y = mx + c

•y=-2x+c

•Substitute ( x, y) = (3,6)

•6=-2(3)+c

•c=12

•The desired equation is:

•y=-2x+12

•

X	0	1	2
Y	8	6	4

X	0	1	2
Y	12	10	8

10. For the line perpendicular to the line y= 2/7 x − 4 and passes through the point (0,9)

•Answer: The line y= 2/7 x − 4

•m= 2/7 , c = - 4

• m (perpendicular) = - 1/m = - 1/(2/7) = - 7/2

partial equation y = - 7/2 x + c

•9 = 0 + c

•C =9

•Thus required equation y = - 7/2 x + 9

•Another method : m = "y -y1" /"x -x1"

•- 7/2 = "y-9" /(" x-" 0)

y - 9=- 7/2(x - 0)

•y - 9 = - 7x/2

•y = - 7x/2 +9

•2(y-9) = -7x

•2y - 18 = -7x

•2y=-7x+18

X	0	2
Y	9	2

X	0	7
Y	-4	-2

11. For the line perpendicular to the line y=1 and passes and passes through the point (3, -10)

•Answer:

•the graph of y=1 is a horizontal line, any line perpendicular to a horizontal line is a vertical line, whose equation can be written in the form x=a. Since it passes through the point (3,-10), so the equation is: x=3

•

m = 0

y-y1 = m (x-x1 )

y +10= (-1)/0 (x - 3)

•0= x-3

•x = 3

X	3
Y	0

X	0
Y	1

12. For the line perpendicular to the line x=7 passes through the point (-9,4)

•Answer:

•The line x = 7 is vertical.

•The line perpendicular is horizontal and passes through the point (-9,4).

•Horizontal lines have slope 0.

•Equation of line: y - 4 = 0(x + 9)

•y = 4

X	0
Y	4

X	7
Y	0

Pages 5 of 42

13. For the line Perpendicular to y-3x=6 and passes through (1, 2)

•Answer: y= 3x+6

•m=3

If two lines are perpendicular, then m1⋅m2=−1

Slope of line perpendicular to y=3x+6 will be (-1)/3

∴ Equation of line y=(-1)/3x+c

•This line passes through the point (1,2)

2=(-1)/3*1 + c

2 + 1/3 = c

c= 7/3

Equation of line becomes x+3y=7

Y = (-1)/3 x + 7/3

•Another method

•y-y1 = m (x-x1 )

•y-2= (-1)/3(x-1)

•Y = (-1)/3 x + 1/3 +2

•Y = (-1)/3 x + 7/3

X	0	1
Y	6	9

X	1	4	7
Y	2	1	0

14. For the line Perpendicular to 2x+3y=6 and passes through (0,-2)

•Answer: y= (-2)/3x+ 2

•m=(-2)/3

If two lines are perpendicular, then m1⋅m2=−1

Slope of line perpendicular to y= (-2)/3x+ 2 will be 3/2

∴ Equation of line y = 3/2x + c

•This line passes through the point (0,-2)

-2=3/2*0 + c

c= -2

Equation of line becomes y = 3/2x - 2

•Another method

•y-y1 = m (x-x1 )

•Y+2= 3/2(x-0)

•Y = 3/2 x - 2

X	0	2	4
Y	-2	1	4

X	0	3
Y	2	0

15. For the line Perpendicular to 2x-4=y and passes through the origin

•Answer: y= 2x - 4

•m=2

If two lines are perpendicular, then m1⋅m2=−1

Slope of line perpendicular to y= 2x - 4 will be (-1)/2

∴ Equation of line y = (-1)/2x + c

•This line passes through the point (0,0)

c= 0

Equation of line becomes y = (-1)/2x

•Another method

•y-y1 = m (x-x1 )

•Y+0= (-1)/2(x-0)

•Y = (-1)/2 x

X	0	2	4
Y	0	-1	-2

X	0	2	4
Y	-4	0	4

16. For the line Perpendicular to 3x+2y=6 and passes through (-2,5)

•Answer: y= (-3)/2x+ 3

•m=(-3)/2

If two lines are perpendicular, then m1⋅m2=−1

Slope of line perpendicular to y= (-3)/2x+ 3 will be 2/3

∴ Equation of line y = 2/3x + c

•This line passes through the point (-2,5)

5=2/3*(-2) + c

c= 5+ 4/3 = 19/3

Equation of line becomes y = 2/3x + 19/3

•Another method

•y-y1 = m (x-x1 )

•Y – 5 = 2/3(x + 2)

•Y = 2/3 x + 4/3+5

•y = 2/3x + 19/3

X	1	4
Y	7	9

X	0	2	4
Y	3	0	-3

Page 6 of 42

17. What is the equation of a line that is perpendicular to the line y=x+7 and passes through the point (-4, 2)?

•Answer: y= x +7

•m = 1

If two lines are perpendicular, then m1⋅m2=−1

Slope of line perpendicular to y= x +7 will be -1

∴ Equation of line y = -1 x + c

•This line passes through the point (-4,2)

2=(-1)*(-4) + c

c= -2

Equation of line becomes y = -x - 2

•Another method

•y-y1 = m (x-x1 )

•Y -2 =-1 (x + 4)

•Y = - x - 4+2

•y = -x -2

X	0	1	2
Y	7	8	9

X	0	1	2
Y	-2	-3	-4

18. What is the equation of a line that is perpendicular to the line 3y=2x-9 and passes through the point (-1, 6)?

X	0	3	6
Y	-3	-1	1

X	1	3	5
Y	3	0	-3

19. For The line through point (-6,4) and is perpendicular to the line y = 1/2 x+2/1

•Answer: y = 1/2 x+2/1

•m =1/2

If two lines are perpendicular, then m1⋅m2=−1

Slope of line perpendicular to y = 1/2 x+2/1 will be -2

∴ Equation of line y =-2 x + c

•This line passes through the point (- 1,6)

4=(-2)*(-6) + c

c= -8

Equation of line becomes y =-2 x - 8

Another method

•y-y1 = m (x-x1 )

•Y - 4 =-2 (x + 6)

•Y = -2x - 12+4

•y =-2 x - 8

X	0	2	4
Y	2	3	4

X	0	-1	-2
Y	-8	-6	-4

20. What is the linear equation through (-2,2) that will create a right angle when it intersects with the line shown in the graph

•Answer: m = "y2 -y1" /"x2 -x1"

•(1,-1), (0,-2)

•m = ("-2" +1)/(" " 0-1) =1

slope perpendicular =−1/m = -1

y-y1 = m (x-x1 )

•Y - 2 =-1 (x + 2)

•Y = -x - 2+2

•y = - x

X	1	0	-1
Y	-1	-2	-3

X	-2	0	1
Y	2	0	-1

Page 7 of 42

G.GPE.B.5: PARALLEL AND PERPENDICULAR LINES

39 Given (MN) ̅ shown below, with M(-6,1) and N(3,-5), what is equation of the line that passes through point P(6,1) and is parallel to (MN) ̅?

Page 7 of 42

In the diagram below ∆ABC has vertices A(4,5), B(2,1) and C(7,3). What is the slope of the altitude drawn from A to (BC) ̅ ?

•slope of BC = (y2 - y1) / (x2 - x1)

slope of BC = (3 - 1) / (7 - 2)

slope of BC = 2/5

•The equation of line BC, B (2, 1) is y - y1 = m(x - x1)

y - 1 = (2/5)(x - 2)

y = (2/5)x - 1/5

•The equation of the line perpendicular to line BC that passes through point A (4, 5).

•slope of altitude = -1 / slope of BC

slope of altitude = -1 / (2/5)

slope of altitude = -5/2

•The equation of the altitude : y - y1 = m(x - x1)

y - 5 = (-5/2)(x - 4)

y = (-5/2)x + 15

•Therefore, the slope of the altitude drawn from A to base BC is -5/2.

•

Page 7 of 42

Which equation represents the line that passes through the point(-2,2) and is parallel to y = 1/2 x +8?

•parallel lines have same slope

•If y = 1/2 x + 8 is parallel then the slope is 1/2

•y = 1/2 x + b

•Substituting the point (-2, 2)

•2 = 1/2 (-2) + b

•2 = -1 + b

•b = 2 + 1

•b = 3

•So the equation is y = 1/2 x + 3

•Therefore, the equation is y = 1/2 x + 3.

Page 8 of 42

Which equation represents a line that is perpendicular to line represented by 2x-y = 7?

•If two lines are perpendicular to each other

•m1 . m2 = -1

•From the equation 2x - y = 7

•Y=2x-7

•m1 = 2

•2 . m2 = -1

•m2 = -1/2

•If the line is perpendicular to Y=2x-7

•Y-0= -1/2 (x -0)

•

•As the y intercept value is not given

•Consider the y intercept as 6

•y = -1/2 x + 6

•Therefore, the equation of the line is y = -1/2 x + 6

•

Page 8 of 42

An equation of a line perpendicular to the line represented by the equation y = "- " 1/2 x - 5 and passing through (6,-4) is

•y = "- " 1/2 x – 5

•m = -1/2

•Slope of the perpendicular line = -1/(-1/2)

•= 1/(1/2) =2

•Then the equation is Y = 2x +c

•Passing through point (6,-4)

•-4 = (2)(6) + c

•-4 = 12 + c

•c = -12 - 4

•c = -16

•y = 2x - 16

•Therefore, the equation of the line is y = 2x - 16.

Page 8 of 42

What is an equation of a line that is perpendicular to the line whose equation is 2y = 3x-10 and passes through (-6,1)?

•2y = 3x-10

•Y = " " 3/2 x -5

•m= 3/2

•New slope m=1/-m="- " 2/3

•passes through the point (-6, 1).

• New equation is; y - 1 = ("- " 2/3)(x + 6)

•y - 1 = "- " 2/3x - 4

•y = "- " 2/3x – 3

•

Page 9 of 42

8B DMS Formulas- Coordinate

Coordinate Geometry

Examples:

1.) Find the slope of the line that passes through the points (5,-6) and (3,-2).

•ANSWER:

• ("x1", "y1") = ( 5,-6) , ("x2","y2") = (3,-2)

•

•m = "y2 -y1" /"x2 -x1"

•m = "-2+6" /" 3 -5"

•m= 4/(-2)

•m= -2

•

2.) Find the midpoint of the segment connecting the points (-2, -9) and (-11, 5).

•Mid point Formula

M = ("x2 + x1" /"2" , "y2 +y 1" /"2" )

M = ("-2 -11" /"2" , "-9+5" /"2" )

M = ((-13)/"2" , ("-" 4)/"2" )

M = (-6.5, -2)

3.) Find the length of the line segment whose endpoints are (-4, -5) and (1, -2)

•("x1", "y1") = ( -4,-5) , ("x2","y2") = (1,-2)

•D = √(〖"(x2" -"x1" )〗^2+〖"(y2" -"y1" )〗^2 )

•D = √(〖"(1" +4)〗^2+〖"(-2" +5)〗^2 )

•D = √(〖"(5" )〗^2+〖"(" 3)〗^2 )

•D = √(25+9)

•D = √34

Page 10 of 42

4.) What is the coordinate of the point directly in the middle of (8, 12) and (6,-3).

To find the coordinates of the point that lies exactly in the middle of two given points, you can use the midpoint formula.

The midpoint formula for a point $(x_{m}, y_{m})$ between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is:

Given the points $(8, 12)$ and $(6, - 3)$ , we can calculate the midpoint as follows:

So, the coordinates of the point directly in the middle of $(8, 12)$ and $(6, - 3)$ are $(7, 4.5)$ .

5.) What is the slope of the line that passes through the points (5, 2) and (-3, 6)?

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(5, 2)$ and $(- 3, 6)$ , you can calculate the change in $y$ and the change in $x$ :

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(5, 2)$ and $(- 3, 6)$ is $- \frac{1}{2}$ .

6.) How far apart are (-3, 6) and (11, 7)?

To find the distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the distance formula:

Given the points $(- 3, 6)$ and $(11, 7)$ , you can calculate the distance as follows:

So, the distance between the points $(- 3, 6)$ and $(11, 7)$ is $\sqrt{197}$ units.

7.) Find the midpoint of the segment connecting the points (18a, 5b) and (-3a, 11b).

To find the midpoint of a line segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the midpoint formula:

In this case, the coordinates of the endpoints are:

- $(x_{1}, y_{1}) = (18 a, 5 b)$

- $(x_{2}, y_{2}) = (- 3 a, 11 b)$

Now, plug these values into the midpoint formula:

So, the midpoint of the segment connecting the points $(18 a, 5 b)$ and $(- 3 a, 11 b)$ is $(\frac{15 a}{2}, 8 b)$ .

8.) Find the slope of the line that passes through the points (-6x,8y) and (2x,-4y)

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(- 6 x, 8 y)$ and $(2 x, - 4 y)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $- 4 y - 8 y = - 12 y$

Change in $x$ = $2 x - (- 6 x) = 2 x + 6 x = 8 x$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(- 6 x, 8 y)$ and $(2 x, - 4 y)$ is $\frac{- 3 y}{2 x}$ .

Page11 of 42
8B DMS Formulas- Coordinate HW Coordinate Geometry HW Slope Formula Distance

Formula Midpoint Formula _________ ___________ _____________

1.) In simplest radical form, how far apart are (-3, 2) and (-7, 4)?

To find the distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the distance formula:

Given the points $(- 3, 2)$ and $(- 7, 4)$ , you can calculate the distance as follows:

Since $20$ is not a perfect square, we can express the distance in simplest radical form as $\sqrt{20}$ units.

2.) Find the slope of the line that passes through the points (6,-2) and (6,5).

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(6, - 2)$ and $(6, 5)$ , you can see that the $x$ -coordinates are the same. Since the line is vertical, the change in $x$ is zero, and division by zero is undefined.

A vertical line does not have a slope in the traditional sense. Instead, we say the slope is "undefined".

So, the slope of the line passing through the points $(6, - 2)$ and $(6, 5)$ is undefined.

3.) Find the midpoint of the segment connecting the points (3x, 9y) and (11x, 7y).

To find the midpoint of a line segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the midpoint formula:

In this case, the coordinates of the endpoints are:

- $(x_{1}, y_{1}) = (3 x, 9 y)$

- $(x_{2}, y_{2}) = (11 x, 7 y)$

Now, plug these values into the midpoint formula:

So, the midpoint of the segment connecting the points $(3 x, 9 y)$ and $(11 x, 7 y)$ is $(7 x, 8 y)$ .

4.) Find the slope of the line that passes through the points (-7,-4) and (8,-4).

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(- 7, - 4)$ and $(8, - 4)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $- 4 - (- 4) = 0$

Change in $x$ = $8 - (- 7) = 8 + 7 = 15$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(- 7, - 4)$ and $(8, - 4)$ is $0$ .

5.) What is the distance between the two given points: (5, 7) and (4, 8)

To find the distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the distance formula:

Given the points $(5, 7)$ and $(4, 8)$ , you can calculate the distance as follows:

So, the distance between the points $(5, 7)$ and $(4, 8)$ is $\sqrt{2}$ units.

Page 12 of 42
6.) Find the slope of the line that passes through the points (2, 9) and (-5, 13).

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(2, 9)$ and $(- 5, 13)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $13 - 9 = 4$

Change in $x$ = $- 5 - 2 = - 7$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(2, 9)$ and $(- 5, 13)$ is $- \frac{4}{7}$ .

Review Section
8.) Two lines are represented by the equations − (−1)/2y = 6x + 10 and y = mx. For which value of m will the lines be parallel?
(1) -12 (2) -3 (3) 3 (4) 12

To determine if two lines are parallel, we need to compare their slopes.

The first equation is in the form $\frac{- 1}{2} y = 6 x + 10$ , which can be rewritten as $y = - 12 x - 20$ .

Comparing it to $y = m x$ , we see that the slope of the first line is $- 12$ .

To find the slope of the second line, we look at the coefficient of $x$ in the second equation, which is $m$ .

For the two lines to be parallel, their slopes must be equal. So, we have:

Therefore, the value of $m$ for which the lines will be parallel is $- 12$ .

9.) The two lines represented by the equations below are graphed on a coordinate plane.
x + 6y = 12
3(x – 2) = -y – 4
Which statement best describes the two lines?
(1) The lines are parallel. (3) The lines are perpendicular
(2) The lines are the same line. (4) The lines intersect at an angle other than 90°

To determine the relationship between the two lines, let's first rewrite the equations in slope-intercept form ( $y = m x + b$ ):

1. $x + 6 y = 12$

Subtract $x$ from both sides:

$6 y = - x + 12$

Divide both sides by 6:

$y = - \frac{1}{6} x + 2$

2. $3 (x - 2) = - y - 4$

Expand and simplify:

$3 x - 6 = - y - 4$

Add 6 to both sides:

$3 x = - y + 2$

Divide both sides by 3:

$y = - \frac{1}{3} x + \frac{2}{3}$

Now, we can compare the slopes of the two lines. The slope of the first line is $- \frac{1}{6}$ , and the slope of the second line is $- \frac{1}{3}$ .

Since the slopes are not equal, the lines are not parallel. To determine if they are perpendicular, we can check if the product of their slopes is -1.

Since the product of the slopes is not -1, the lines are not perpendicular either.

Hence, the correct statement is:

(4) The lines intersect at an angle other than 90°

10.) In ∆ABC, BD is an perpendicular bisector. If AD = 2y + 4 and CD = y + 12 and m∠BDA = 5(x – 12), find the value of x and y.

Since BD is the perpendicular bisector of segment AC, it divides AC into two equal parts. Therefore, AD = CD.

From the given information:

AD = 2y + 4

CD = y + 12

Setting these equal to each other:

2y + 4 = y + 12

Subtracting y from both sides:

y + 4 = 12

Subtracting 4 from both sides:

y = 8

Now that we found the value of y, we can substitute it into one of the expressions for AD or CD to find its value. Let's use AD:

AD = 2y + 4

AD = 2(8) + 4

AD = 16 + 4

AD = 20

Now, we can use the given information about angle BDA to find the value of x:

m∠BDA = 5(x – 12)

Given that BD is a perpendicular bisector, angle BDA will be a right angle. Therefore, m∠BDA = 90°.

Substitute 90° for m∠BDA:

90 = 5(x – 12)

Now, solve for x:

90 = 5x - 60

150 = 5x

x = 30

So, the value of x is 30 and the value of y is 8.

.11) Using the given diagram, create a two column proof to prove that m ∠ w+ m ∠ x + m ∠ z = 180°

**Statement** | **Reason**

--- | ---

m∠w = m∠w | Identity property of equality

m∠x = m∠x | Identity property of equality

m∠z = m∠z | Identity property of equality

m∠w + m∠x + m∠z = m∠w + m∠x + m∠z | Addition property of equality

m∠w + m∠x + m∠z = 180° | Definition of straight angle (m∠w, m∠x, and m∠z form a straight angle, which measures 180°)

Examples:

1.) Find the slope of the line that passes through the points (-2, 7) and (4, 5).

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(- 2, 7)$ and $(4, 5)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $5 - 7 = - 2$

Change in $x$ = $4 - (- 2) = 4 + 2 = 6$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(- 2, 7)$ and $(4, 5)$ is $- \frac{1}{3}$ .

2. Line segment AB has endpoints A(2, -3) and B(-4 6). What are the coordinates of the midpoint of (𝐴𝐵) ̅?

To find the coordinates of the midpoint of a line segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the midpoint formula:

In this case, the coordinates of the endpoints are:

- $(x_{1}, y_{1}) = (2, - 3)$ (point A)

- $(x_{2}, y_{2}) = (- 4, 6)$ (point B)

Now, plug these values into the midpoint formula:

So, the coordinates of the midpoint of the line segment AB with endpoints A(2, -3) and B(-4, 6) are $(- 1, \frac{3}{2})$ .

3.) Find the length of the line segment whose endpoints are (-4, -9) and (2, -5)

To find the length of the line segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the distance formula:

In this case, the coordinates of the endpoints are:

- $(x_{1}, y_{1}) = (- 4, - 9)$

- $(x_{2}, y_{2}) = (2, - 5)$

Now, plug these values into the distance formula:

So, the length of the line segment with endpoints (-4, -9) and (2, -5) is $\sqrt{52}$ units.

4.) What are the coordinates of the center of a circle if the endpoints of its diameter are A(6, 10) and B(-8, -4).

To find the center of the circle given the endpoints of its diameter, we first find the midpoint of the diameter. The midpoint of a line segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:

Given the endpoints $A (6, 10)$ and $B (- 8, - 4)$ , we can find the midpoint:

So, the coordinates of the midpoint, which is also the center of the circle, are $(- 1, 3)$ .

Page 15 of 42
5.) What is the slope of the line that passes through the points (-3x, 6y) and (1x, 9y)?

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(- 3 x, 6 y)$ and $(1 x, 9 y)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $9 y - 6 y = 3 y$

Change in $x$ = $1 x - (- 3 x) = 1 x + 3 x = 4 x$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(- 3 x, 6 y)$ and $(1 x, 9 y)$ is $\frac{3 y}{4 x}$ .

6.) Given circle O with diameter AB, A(4, 9) and B(-2, 13), what is the length of the diameter?

To find the length of the diameter of a circle given its endpoints, you can use the distance formula.

The distance formula between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is:

Given the endpoints A(4, 9) and B(-2, 13) of the diameter, you can calculate the distance between them:

So, the length of the diameter of the circle is $\sqrt{52}$ units.

7.) The line connecting the points (1, 6) and (-5, 14) has a slope of what?

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(1, 6)$ and $(- 5, 14)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $14 - 6 = 8$

Change in $x$ = $- 5 - 1 = - 6$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(1, 6)$ and $(- 5, 14)$ is $- \frac{4}{3}$ .

8.) If a line segment has endpoints A(3x + 5, 3y) and B( x – 1, -7y), what are the coordinates of the midpoint of AB?

To find the midpoint of a line segment with endpoints $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ , you can use the midpoint formula:

In this case, the coordinates of the endpoints are:

- $A (3 x + 5, 3 y)$

- $B (x - 1, - 7 y)$

Now, plug these values into the midpoint formula:

Simplify the expression:

So, the coordinates of the midpoint of the line segment AB are $(2 x + 2, - 2 y)$ .

9.) In the provided diagram, what is the slope of the line passing through points A and B?

•A(-3,4)

•B(5,8)

•(Y2-Y1)/(X2-X1) = (8-4)/(5+3) = 4/8

Page 16 of 42
Find an endpoint given one endpoint and the Midpoint

10.) M is the midpoint of segment CD. The coordinates of M(-1,1) and C(1,3) are given. Find the coordinates of point D.

To find the coordinates of point D, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the midpoint M is $(- 1, 1)$ and point C is $(1, 3)$ , we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of point D:

1. Solve for the x-coordinate:

2. Solve for the y-coordinate:

So, the coordinates of point D are $(- 3, - 1)$ .

11.) M is the midpoint of segment AB. The coordinates of M(-4,5) and B(10,7) are given. Find the coordinates of point A

To find the coordinates of point A, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the midpoint M is $(- 4, 5)$ and point B is $(10, 7)$ , we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of point A:

1. Solve for the x-coordinate:

2. Solve for the y-coordinate:

So, the coordinates of point A are $(- 18, 3)$ .

12.) The coordinates of M, the midpoint of segment JA are (-91,-8). J(-120,-9) and A(x,y)

To find the coordinates of point A, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the midpoint M is $(- 91, - 8)$ and point J is $(- 120, - 9)$ , we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of point A:

1. Solve for the x-coordinate:

2. Solve for the y-coordinate:

So, the coordinates of point A are $(- 62, - 7)$ .

13.) If the coordinates of the midpoint of segment AB are (8,2) and A(-9,-6) find the coordinates of point B

To find the coordinates of point B, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the midpoint M is $(8, 2)$ and point A is $(- 9, - 6)$ , we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of point B:

1. Solve for the x-coordinate:

2. Solve for the y-coordinate:

So, the coordinates of point B are $(25, 10)$ .

Page 17 of 42
8C DMS Formulas- Coordinate Day 2 HW Coordinate Geometry (Day 2) HW Slope Formula Distance Formula Midpoint Formula __________

1.) The coordinates of the endpoints of FG are (-4,3) and (2,5). Find the length of FG in simplest radical form

To find the length of the line segment FG with endpoints $(- 4, 3)$ and $(2, 5)$ , we can use the distance formula:

Given the endpoints $(- 4, 3)$ and $(2, 5)$ , we can plug these values into the distance formula:

So, the length of the line segment FG in simplest radical form is $\sqrt{40}$ units.

2.) In the diagram below, quadrilateral ABCD has vertices A(-5,1) B(6,-1) C(3,5) and D(-2,7). What are the coordinates of the midpoint of diagonal AC?

To find the midpoint of diagonal AC, we'll use the midpoint formula. The midpoint formula states that the coordinates of the midpoint between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the endpoints of diagonal AC are A(-5, 1) and C(3, 5), we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of the midpoint:

So, the coordinates of the midpoint of diagonal AC are $(- 1, 3)$ .

3.) In the provided diagram, what is the slope of the line passing through points A and B?

•A(-3,7)

•B(3,6)

•(Y2-Y1)/(X2-X1) = (6-7)/(3+3) = (-1)/6

4.) If M is the midpoint of PQ,P = 5, M = 14, Q=x. find the value of x

If $M$ is the midpoint of $P Q$ , then the coordinates of $M$ will be the average of the coordinates of $P$ and $Q$ since $M$ is exactly halfway between $P$ and $Q$ .

So, if $P = 5$ and $M = 14$ , the average of $P$ and $Q$ should equal $M$ .

To solve for $x$ , we can multiply both sides of the equation by $2$ :

Subtracting $5$ from both sides:

So, $x = 23$ .

Page 18 of 42
5.) Find the slope of the line that passes through the points (-4a, 8b) and (6a, 4b).

To find the slope of the line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , you can use the formula for slope:

Given the points $(- 4 a, 8 b)$ and $(6 a, 4 b)$ , you can calculate the change in $y$ and the change in $x$ :

Change in $y$ = $4 b - 8 b = - 4 b$

Change in $x$ = $6 a - (- 4 a) = 6 a + 4 a = 10 a$

Now, plug these values into the formula for slope:

So, the slope of the line passing through the points $(- 4 a, 8 b)$ and $(6 a, 4 b)$ is $- \frac{2 b}{5 a}$ .

6.) M is the midpoint of segment TA. M(0, 8) and T(9, 9) find the coordinates of A.

To find the coordinates of point A, we can use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the midpoint M is $(0, 8)$ and point T is $(9, 9)$ , we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of point A:

1. Solve for the x-coordinate:

2. Solve for the y-coordinate:

So, the coordinates of point A are $(- 9, 7)$ .

7.) The coordinates of M, the midpoint of segment TY are (-54,65) and Y(-58,62). What are the coordinates of G?

To find the coordinates of point G, we'll use the midpoint formula. The midpoint formula states that the coordinates of the midpoint M between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ are:

Given that the midpoint M is $(- 54, 65)$ and point Y is $(- 58, 62)$ , we can plug these values into the midpoint formula:

Now, let's solve for the coordinates of point G:

1. Solve for the x-coordinate:

2. Solve for the y-coordinate:

So, the coordinates of point G are $(- 50, 68)$ .

8.) Jerry and Jean Jogger start at the same time from point A sown on te accompanying set of aces. Jerry jogs at a rate of 5 miles per hour traveling from point A to point R to point S and then to point C. Jean jogs directly from point A to point C at a rate of 3 miles per hour. Which jogger reaches point C first? Explain or show your reasoning. Remember: D=RT

To determine who reaches point C first, we can calculate the time it takes for each jogger to reach point C.

Let's start with Jerry:

1. First, Jerry jogs from point A to point R.

2. Then, he jogs from point R to point S.

3. Finally, he jogs from point S to point C.

Given that the distances from A to R, R to S, and S to C are not specified, we can represent them as $d_{1}$ , $d_{2}$ , and $d_{3}$ respectively.

For Jerry:

- Time to reach R: $t_{1} = \frac{d_{1}}{5}$

- Time to reach S from R: $t_{2} = \frac{d_{2}}{5}$

- Time to reach C from S: $t_{3} = \frac{d_{3}}{5}$

Total time for Jerry to reach C: $T_{J} = t_{1} + t_{2} + t_{3}$

Now, let's calculate the total time for Jean to reach C:

Since Jean jogs directly from A to C, the total distance she travels is $d_{4}$ , the distance from A to C.

For Jean:

- Time to reach C: $T_{J} = \frac{d_{4}}{3}$

If $T_{J}$ for Jerry is less than $T_{J}$ for Jean, then Jerry reaches C first. Otherwise, Jean reaches C first.

Therefore, to determine who reaches point C first, we need to know the distances from A to R, R to S, and S to C, or alternatively, the total distance from A to C for comparison. Without this information, we cannot definitively determine who reaches point C first.

Page 19 of 42
Review Section: _____
9.) What is an equation of the line that passes through the point (-2,5) and is perpendicular to the line whose equation is Y= 1/2 𝑋 + 5? (1) y = 2x + 1 (2) y = -2x + 1 (3) y = 2x + 9 (4) y = -2x – 9

To find the equation of the line perpendicular to $y = \frac{1}{2} x + 5$ that passes through the point $(- 2, 5)$ , we need to determine the slope of the perpendicular line.

The slope of the given line $y = \frac{1}{2} x + 5$ is $\frac{1}{2}$ .

Perpendicular lines have slopes that are negative reciprocals of each other. So, the slope of the line perpendicular to $y = \frac{1}{2} x + 5$ will be the negative reciprocal of $\frac{1}{2}$ , which is $- 2$ .

Now, we have the slope ( $m = - 2$ ) and a point $(- 2, 5)$ . We can use the point-slope form of a linear equation to find the equation of the perpendicular line:

where $m$ is the slope and $(x_{1}, y_{1})$ is the given point.

Substitute the values $m = - 2$ , $x_{1} = - 2$ , and $y_{1} = 5$ :

So, the equation of the line perpendicular to $y = \frac{1}{2} x + 5$ and passing through the point $(- 2, 5)$ is $y = - 2 x + 1$ .

______10.) Line k is drawn so that it is perpendicular to two distinct planes, P and R. What must be true about planes P and R? (1) Planes P and R are skew (2) Planes P and R are parallel (3) Planes P and R are perpendicular (4) Plane P intersects Plane R but is not perpendicular

•When a line is perpendicular to two distinct planes, those planes must be parallel to each other. This is because a line can be perpendicular to only one plane at a time. If a line is perpendicular to two planes, it must be perpendicular to every line in the plane that is perpendicular to it. This means that the planes must be parallel to each other.

•

•Therefore, the correct answer is:

•

•(2) Planes P and R are parallel

______11.) Which of the following sets of numbers could represent the sides of a triangle? (1) {3,4,7} (2) {8,10,19} (3) {2,3,4} (4) {1,1,3}

•To determine if a set of numbers could represent the sides of a triangle, we need to check if the sum of the lengths of any two sides is greater than the length of the third side for all combinations of sides.

Let's go through each set of numbers:

1. {3, 4, 7}

- Sum of any two sides: 3 + 4 = 7, 4 + 7 = 11, 3 + 7 = 10

- In this case, the sum of the lengths of two sides (3 and 4) is greater than the length of the third side (7), so this set could represent the sides of a triangle.

2. {8, 10, 19}

- Sum of any two sides: 8 + 10 = 18, 10 + 19 = 29, 8 + 19 = 27

- The sum of the lengths of the two smaller sides (8 and 10) is less than the length of the third side (19), so this set cannot represent the sides of a triangle.

3. {2, 3, 4}

- Sum of any two sides: 2 + 3 = 5, 3 + 4 = 7, 2 + 4 = 6

- The sum of the lengths of any two sides is always greater than the length of the third side, so this set could represent the sides of a triangle.

4. {1, 1, 3}

- Sum of any two sides: 1 + 1 = 2, 1 + 3 = 4, 1 + 3 = 4

- The sum of the lengths of any two sides is not greater than the length of the third side for all combinations, so this set cannot represent the sides of a triangle.

Therefore, the sets of numbers that could represent the sides of a triangle are (1) {3, 4, 7} and (3) {2, 3, 4}.

12.) Trapezoid TRAP, with median MQ, is shown in the diagram below. TR = 12Y+1, MQ = 16Y+1 AND PA = 18Y+6. Solve algebraically for the value of y

•To solve for the value of $y$ , we'll use the properties of a trapezoid with a median.

In a trapezoid with median, the median divides the trapezoid into two congruent triangles. This means that the lengths of the segments on either side of the median (from the midpoint of the non-parallel sides to the endpoints of the parallel sides) are equal.

Given:

1. $T R = 12 y + 1$

2. $M Q = 16 y + 1$

3. $P A = 18 y + 6$

Since MQ is a median, it divides TRAP into two congruent triangles. So, $T R = P A$ .

Equating TR and PA:

Now, solve for $y$ :

So, the value of $y$ is $- \frac{5}{6}$ .

Page 20 of 42

Page 21 of 42

8D DMS Formulas- Coordinate Day 3 Distance, Midpoint, Slope Day 3 Distance Formula:

1.) Distance: The distance formula can be found by using the Pythagorean Theorem. Given two points (x1, y1) and (x2, y2) the distance formula is: distance = __________________________

•The distance formula, derived from the Pythagorean Theorem, gives the distance between two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in a Cartesian coordinate system. It's represented as:

This formula calculates the length of the line segment connecting the two points by taking the square root of the sum of the squares of the differences in their $x$ and $y$ coordinates.

2.) Midpoint: A point halfway between the endpoints of a line segment is called the midpoint. A midpoint divides a segment into two equal segments. midpoint =

•The midpoint of a line segment with endpoints $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in a Cartesian coordinate system can be found using the midpoint formula:

This formula calculates the average of the $x$ coordinates of the endpoints to find the $x$ coordinate of the midpoint, and similarly for the $y$ coordinate.

3.) Slope: The slope of a line is a rate of change and is represented by m. In coordinate geometry, the formula for slope given the two points (x1, y1) and (x2, y2). slope = ______________________________ Line Relationships with Slope: Parallel Lines: Lines that never intersect. They have the ___________ slope

Perpendicular Lines: Lines that meet to form right angles. These lines have _______________________ slopes

•The formula for calculating the slope $m$ of a line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:

For line relationships with slope:

- Parallel lines have the same slope.

- Perpendicular lines have slopes that are negative reciprocals of each other.

So, for parallel lines, they have the ___________ slope, and for perpendicular lines, they have _______________________ slopes.

Page 22 of 42

4.) Given A(6,2) and B(8,-4), and C(9,3) and D(6,2), how are lines AB and CD related? (Hint: are they parallel, perpendicular, or neither?).

•To determine how the lines $A B$ and $C D$ are related, we need to find the slopes of both lines.

For the line $A B$ passing through points $A (6, 2)$ and $B (8, - 4)$ , the slope $m_{A B}$ is calculated using the slope formula:

Substituting the coordinates of points $A$ and $B$ :

For the line $C D$ passing through points $C (9, 3)$ and $D (6, 2)$ , the slope $m_{C D}$ is calculated similarly:

Since the slopes of $A B$ and $C D$ are not equal (they are not the same value), the lines are not parallel. To determine if they are perpendicular, we need to check if the product of their slopes is -1.

The product of their slopes is -1, indicating that the lines $A B$ and $C D$ are perpendicular to each other.

5.) Given J(-8,-2) and K(-5,-6), and L(9,-7) and M(5,-10), how are lines JK and LM related?

•To determine how the lines $J K$ and $L M$ are related, we need to find the slopes of both lines.

For the line $J K$ passing through points $J (- 8, - 2)$ and $K (- 5, - 6)$ , the slope $m_{J K}$ is calculated using the slope formula:

Substituting the coordinates of points $J$ and $K$ :

For the line $L M$ passing through points $L (9, - 7)$ and $M (5, - 10)$ , the slope $m_{L M}$ is calculated similarly:

Since the slopes of $J K$ and $L M$ are not equal (they are not the same value), the lines are not parallel. To determine if they are perpendicular, we need to check if the product of their slopes is -1.

The product of their slopes is -1, indicating that the lines $J K$ and $L M$ are perpendicular to each other.

6.) Line m and point P are shown in the graph below. Which equation represents the line passing through P and parallel to line m? (1) Y − 3 = 2(X + 2) (2) Y + 2 = 2(X − 3) (3) Y − 3 = − 1/2(X + 2) (4) Y + 2 = − 1/2(X − 3)

•First, let's determine the slope of line $m$ . The given equation of line $m$ is $y = 2 x + 3$ , which is already in slope-intercept form ( $y = m x + b$ ). In this form, $m$ represents the slope.

Comparing the equation $y = 2 x + 3$ with the slope-intercept form, we see that the slope $m$ is $2$ .

Since the line parallel to $m$ will have the same slope, we can use the point-slope form to find its equation. The point-slope form is:

Given that point $P (3, - 2)$ and $m = 2$ , we substitute these values into the point-slope form:

Therefore, the correct equation representing the line passing through $P$ and parallel to line $m$ is:

Page 23 of 42

8D DMS Formulas- Coordinate Day 3 HW Distance, Midpoint, Slope Day 3: Homework Complete the questions below. Show all work, when necessary.

•1.) Given points A(2,2), B(10,16), C(-3,1), and D(4,-3), are segments AB and CD perpendicular? Are the lines containing the segments perpendicular? Explain.

To determine if line segments $A B$ and $C D$ are perpendicular, we need to check if the slopes of the lines containing these segments are negative reciprocals of each other.

The slope $m$ of a line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by the formula:

For segment $A B$ , with points $A (2, 2)$ and $B (10, 16)$ :

For segment $C D$ , with points $C (- 3, 1)$ and $D (4, - 3)$ :

Now, we check if the product of their slopes is -1:

Since the product of the slopes is -1, we can conclude that the lines containing segments $A B$ and $C D$ are perpendicular.

2.) Given points N(7,6) and M(7,-2): a.) Write the equation of the line through M and perpendicular to MN. b.) Write the equation of the line through N and perpendicular to MN.

To find the equations of lines perpendicular to $M N$ passing through points $M (7, - 2)$ and $N (7, 6)$ , we need to consider that the slope of a line perpendicular to $M N$ will be the negative reciprocal of the slope of $M N$ .

Given points $N (7, 6)$ and $M (7, - 2)$ , we can first find the slope of $M N$ :

The slope of $M N$ is undefined because the line is vertical.

a.) For the line passing through $M$ and perpendicular to $M N$ , it will be a horizontal line with equation $y = b$ , where $b$ is the y-coordinate of point $M$ :

b.) For the line passing through $N$ and perpendicular to $M N$ , it will be a vertical line with equation $x = a$ , where $a$ is the x-coordinate of point $N$ :

So, the equations of the lines are:

a.) $y = - 2$

b.) $x = 7$

3.) The coordinates of M, the midpoint of segment JA are (-91,-8). J(-120,-9) and A(x,y) find the values of x & y

•To find the coordinates of point $A (x, y)$ , knowing that $M$ is the midpoint of segment $J A$ and the coordinates of $M$ are $(- 91, - 8)$ , and the coordinates of $J$ are $(- 120, - 9)$ , we can use the midpoint formula.

•The midpoint formula states that if the coordinates of the endpoints of a line segment are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , then the coordinates of the midpoint $M$ are given by:

•

•Given that $M (- 91, - 8)$ and $J (- 120, - 9)$ , we can plug these values into the midpoint formula:

•

•Now, solve these equations to find the values of $x$ and $y$ :

•

•So, the coordinates of point $A$ are $(- 62, - 7)$ .

4.) If the coordinates of the midpoint of segment AB are (8,2) and A(-9,-6) find the coordinates of point B.

To find the coordinates of point $B$ , knowing that the midpoint of segment $A B$ is $(8, 2)$ and the coordinates of $A$ are $(- 9, - 6)$ , we can use the midpoint formula.

The midpoint formula states that if the coordinates of the endpoints of a line segment are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , then the coordinates of the midpoint $M$ are given by:

Given that the midpoint $M$ is $(8, 2)$ and the coordinates of $A$ are $(- 9, - 6)$ , we can plug these values into the midpoint formula:

So, the coordinates of point $B$ are $(25, 10)$ .

Page 24 of 42

5.) The coordinates of M, the midpoint of segment TY are (-54,65) and Y(-58,62). What are the coordinates of G?

To find the coordinates of point $G$ , we can use the midpoint formula. The midpoint formula states that if the coordinates of the endpoints of a line segment are $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ , then the coordinates of the midpoint $M$ are given by:

Given that the midpoint $M$ is $(- 54, 65)$ and the coordinates of $Y$ are $(- 58, 62)$ , we can plug these values into the midpoint formula:

So, the coordinates of point $G$ are $(- 50, 68)$ .

6.) Given A(-2,1) and B(1,2), how is line CD related to line AB when C(2,1) and D(5,2). Prove & explain

To determine how line $C D$ is related to line $A B$ , we need to analyze their slopes.

The slope of a line passing through two points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by the formula:

For line $A B$ , with points $A (- 2, 1)$ and $B (1, 2)$ , the slope is:

Now, let's find the slope of line $C D$ , with points $C (2, 1)$ and $D (5, 2)$ :

Since the slopes of both lines $A B$ and $C D$ are the same ( $\frac{1}{3}$ ), we can conclude that lines $A B$ and $C D$ are parallel.

This result can also be visually confirmed by observing that both lines have the same slope and therefore have the same steepness, even though they are at different positions on the coordinate plane. Therefore, the lines are parallel.

Review Questions:
7.) The second side of a triangle is two more than the first side, and the third side is three less than the first side. Which expression represents the perimeter of the triangle? (1) x + 5 (2) 2x – 1 (3) 3x – 1 (4) x² – x – 6

Let's denote the first side of the triangle as $x$ .

According to the given conditions:

- The second side is two more than the first side, so its length is $x + 2$ .

- The third side is three less than the first side, so its length is $x - 3$ .

The perimeter of a triangle is the sum of the lengths of all its sides. Therefore, the perimeter $P$ of the triangle can be expressed as:

Simplifying this expression, we get:

So, the expression $3 x - 1$ represents the perimeter of the triangle. Therefore, the correct answer is option (3) $3 x - 1$ .

8.) The measures of two complementary angles are represented by (3x + 15) and (2x – 10). What is the value of x? (1) 17 (2) 19 (3) 35 (4) 37

Complementary angles add up to $90^{\circ}$ . Therefore, we can set up the equation:

Combine like terms:

Subtract 5 from both sides:

Divide both sides by 5:

So, the value of $x$ is 17. Therefore, the correct answer is option (1) 17.

9.) On the banks of a river, surveyors marked locations A, B, and C. The measure of ∠ACB=70° and the measure of ∠ABC=65° which expression shows the relationship between the lengths of the sides of this triangle?
(1) AB < BC < AC
(2) BC < AB < AC
(3) BC < AC < AB
(4) AC < AB < BC
Based on the given angle measures, we can infer that the side opposite angle $∠ A C B$ (side $A B$ ) is shorter than the side opposite angle $∠ A B C$ (side $B C$ ), which is shorter than the side opposite angle $∠ B A C$ (side $A C$ ).

So, the correct expression showing the relationship between the lengths of the sides of the triangle is:

Thus, the correct answer is option (3) $B C < A C < A B$ .

Page 25 of 42
8E DMS Formulas- Perpendicular Bisector Perpendicular Bisectors: A perpendicular bisector is a line that bisects (divides in half) a segment and that is perpendicular (forms a right angle) to a segment. Therefore, in order to find the perpendicular bisector, we have to find the midpoint of the segment and the slope; then use the point-slope formula to find the equation of the perpendicular bisector.

Examples: 1.) Write an equation of the perpendicular bisector of the line segment whose endpoints are (-1,1) and (7,-5). Graph the original segment and the perpendicular bisector.

To find the equation of the perpendicular bisector of the line segment with endpoints (-1, 1) and (7, -5), we first need to find the midpoint of the segment, which will lie on the perpendicular bisector. Then, we can find the slope of the original line segment and use the negative reciprocal of that slope to find the slope of the perpendicular bisector. Finally, we can use the midpoint and the slope to write the equation of the perpendicular bisector.

1. **Midpoint**:

The midpoint formula is given by: