Question

(Please help, problem is in the photo.)

Answer 1

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

Answer 2

`\sf y = (6)/(5) x-8`

Step-by-step explanation:

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

slope:

`\sf (y-y1)/(x-x1)`

`\hookrightarrow \ \sf (-2-3)/(5--1)`

`\hookrightarrow \ \sf (-5)/(6)`

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

`\sf -(m)^(-1)`

`\hookrightarrow \ \sf -((-5)/(6))^(-1)`

`\hookrightarrow \ \ \sf (6)/(5)`

make equation using:

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

`\hookrightarrow \ \sf y - -2 = (6)/(5) (x-5)`

`\hookrightarrow \ \sf y +2 = (6)/(5) x-6`

`\hookrightarrow \ \sf y = (6)/(5) x-6-2`

`\hookrightarrow \sf y = (6)/(5) x-8`