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lilavasa [31]
3 years ago
12

(Please help, problem is in the photo.)

Mathematics
2 answers:
oksano4ka [1.4K]3 years ago
7 0

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

Rama09 [41]3 years ago
6 0
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
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