Question

(Please help, problem is in the photo.)

Answer 1

Answer:

$\sf y = \dfrac{6}{5} x-8$

explanation:

coordinates: (-1, 3), (5, -2)

slope:

$\sf \dfrac{y-y1}{x-x1}$

$\hookrightarrow \ \sf \dfrac{-2-3}{5--1}$

$\hookrightarrow \ \sf \dfrac{-5}{6}$

The line L is perpendicular to this slope. so the slope will be:

$\sf -(m)^{-1}$

$\hookrightarrow \ \sf -(\dfrac{-5}{6})^{-1}$

$\hookrightarrow \ \ \sf \dfrac{6}{5}$

make equation using:

$\sf y - y1 = m(x-x1)$

$\hookrightarrow \ \sf y - -2 = \dfrac{6}{5} (x-5)$

$\hookrightarrow \ \sf y +2 = \dfrac{6}{5} x-6$

$\hookrightarrow \ \sf y = \dfrac{6}{5} x-6-2$

$\hookrightarrow \sf y = \dfrac{6}{5} x-8$

Answer 2

First we need to find the gradient of K
which is y1-y2/x1-x2
(-1,3) and (5,-2)
so it becomes 3-(-2)/-1-5
m=-5/6
when two lines are perpendicular their gradients multiply to make -1
that means the gradient of L has to be 6/5
we can substitute the point on L (5,-2) and the gradient of 6/5 into y=mx+c
-2 = (6/5) x 5 + c
c = -8
the equation of line L is y= 6x/5 -8