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Question

1. Point P is on segment AB such that AP: PB is 2:5. If A has coordinates (-8, -2) and B has coordinates (6, 19), determine and state the coordinates of P.
2. Point P is on segment AB such that AP: PB is 1:3. If A has coordinates (-3, -4) and B has coordinates (5, 0), determine and state the coordinates of P.
3. Line v has an equation of y = –3/8 x + 5. Perpendicular to line v is line w, which passes through the point (6, 6). What is the equation of line w?
4. Line a has an equation of y = –1/3 x + 6. Line b includes the point (–1, 2) and is parallel to line a. What is the equation of line b?
5. The coordinates of the endpoints of AB are A(3,6) and B(7,–10). Point C is on AB and divides it such that AC:BC is 1:3. What are the coordinates of C?

Answer 1

1. Point P is on segment AB such that AP: PB is 2:5. If A has coordinates (-8, -2) and B has coordinates (6, 19), determine and state the coordinates of P.

P is 2/5 of the way from A to B.

compute the distance from the x-coordinate of B to the x coordinate of A = 6--8 = 14.

Multiply by 2/5 = 5.6. And add that to the x-coordinate of A = 5.6 + -8 = -2.4

compute the distance from the y-coordinate of B to the y coordinate of A = 19--2 = 21.

Multiply by 2/5 = 8.4. And add to the y-coordinate of A = 8.4 + -2 = 6.4.

Coordinates of P are (-4.4,6.4)

1. Point P is on segment AB such that AP: PB is 1:3. If A has coordinates (-3, -4) and B has coordinates (5, 0), determine and state the coordinates of P.

P is 1/4 of the way from A to B.

compute the distance from the x-coordinate of B to the x coordinate of A = 5--3 = 8.

Multiply by 1/4 = 2. And add that to the x-coordinate of A = 2 + -3 = -1

compute the distance from the y-coordinate of B to the y coordinate of A = 0--4 = 4.

Multiply by 1/4 = 1. And add to the y-coordinate of A = 1 + -4 = -3.

Coordinates of P are (-1,3)

1. Line v has an equation of y = –3/8 x + 5. Perpendicular to line v is line w, which passes through the point (6, 6). What is the equation of line w?

slope of line v is -3/8. slope of line w is negative reciprocal of slope of line v = 8/3

y = 8/3x + b. Plug in (6,6) for (x,y) in that equation to solve for b:

6 = 8/3(6) + b

b= -10

y = 8/3x -10

1. Line a has an equation of y = –1/3 x + 6. Line b includes the point (–1, 2) and is parallel to line a. What is the equation of line b?

slope of line a = -1/3. slope of line b parallel to line a is identical = -1/3

y = -1/3x + b. Plug in (-1,2) for (x,y) in that equation to solve for b:

2 = -1/3(-1) + b

5/3 = b

y = -1/3x + 5/3

1. The coordinates of the endpoints of AB are A(3,6) and B(7,–10). Point C is on AB and divides it such that AC:BC is 1:3. What are the coordinates of C?

Follow procedure done for 2nd problem which also has segments in ratio 1:3