Alfredeim
2022-09-11
Answered

Find the equation of the line perpendicular to $y=-\frac{2}{5}x-1$ at x=−1

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Griffin Dean

Answered 2022-09-12
Author has **13** answers

given that $y=-\frac{2}{5}x-1$

at x=−1

$y=-\frac{2}{5}(-1)-1=\frac{2}{5}-1=-\frac{3}{5}$

the line perpendicular has a slope, m where

$m\cdot (-\frac{2}{5})=-1$ $\to m=\frac{5}{2}$

the equation is,

y=mx+c and plug in the values of x, y and m to find c

$-\frac{3}{5}=\frac{5}{2}\cdot (-1)+c$

$-\frac{3}{5}=-\frac{5}{2}+c$

$-\frac{3}{5}+\frac{5}{2}=c$$\to \frac{19}{10}$

therefore the equation is $y=\frac{5}{2}x+\frac{19}{10}$ $\to 10y=25x+19$

at x=−1

$y=-\frac{2}{5}(-1)-1=\frac{2}{5}-1=-\frac{3}{5}$

the line perpendicular has a slope, m where

$m\cdot (-\frac{2}{5})=-1$ $\to m=\frac{5}{2}$

the equation is,

y=mx+c and plug in the values of x, y and m to find c

$-\frac{3}{5}=\frac{5}{2}\cdot (-1)+c$

$-\frac{3}{5}=-\frac{5}{2}+c$

$-\frac{3}{5}+\frac{5}{2}=c$$\to \frac{19}{10}$

therefore the equation is $y=\frac{5}{2}x+\frac{19}{10}$ $\to 10y=25x+19$

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I know I want to be using Gaussian Elimination here, I've augmented the matrix and I'm perfectly familiar with ERO's and back-solving for systems without unknown constants but this is new to me.

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For

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