A few days ago
Anonymous

Pre calculas help!?

find and equation for the line satisfying the given conditions.

1.x intercept= -5 y-intercept= -5

2. through (-1,3) and perpendicular to the line through (0,1) and (4,3)

3. y-intercept 3 and perependicular to 2x-y+6=0

4. find the real number k such that 3,-2 is on the line kx-2y+7=0

Top 1 Answers
A few days ago
cjcourt

Favorite Answer

1. x-intercept = (-5, 0)

y-intercept = (0, -5)

slope = m = rise/run = (y2 – y1) / (x2 – x1) = -5/5 = -1

Put the line in y = mx + b format.

y – y1 = m(x – x1)

y – 0 = -1(x + 5)

y = -x – 5

2. slope = (3 – 1)/(4 – 0) = 2/4 = 1/2

The perpendicular line’s slope is the negative reciprocal of that, i.e., -2.

y – 3 = -2(x + 1)

y = -2x + 1

3. y-intercept (0, 3)

-y = -2x – 6

y = 2x + 6

m = 2 so the negative reciprocal is -1/2

y – 3 = -(1/2)(x – 0)

y = -x/2 + 3

4. Plug in x and y and solve for k.

k(3) – 2(-2) + 7 = 0

3k + 11 = 0

k = -11/3

0