dalasiafarrington
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(Please help, problem is in the photo.)

Please Help Problem Is In The Photo class=

Sagot :

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
fieryanswererft

Answer:

[tex]\sf y = \dfrac{6}{5} x-8[/tex]

explanation:

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

slope:

[tex]\sf \dfrac{y-y1}{x-x1}[/tex]

[tex]\hookrightarrow \ \sf \dfrac{-2-3}{5--1}[/tex]

[tex]\hookrightarrow \ \sf \dfrac{-5}{6}[/tex]

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

[tex]\sf -(m)^{-1}[/tex]

[tex]\hookrightarrow \ \sf -(\dfrac{-5}{6})^{-1}[/tex]

[tex]\hookrightarrow \ \ \sf \dfrac{6}{5}[/tex]

make equation using:

[tex]\sf y - y1 = m(x-x1)[/tex]

[tex]\hookrightarrow \ \sf y - -2 = \dfrac{6}{5} (x-5)[/tex]

[tex]\hookrightarrow \ \sf y +2 = \dfrac{6}{5} x-6[/tex]

[tex]\hookrightarrow \ \sf y = \dfrac{6}{5} x-6-2[/tex]

[tex]\hookrightarrow \sf y = \dfrac{6}{5} x-8[/tex]

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