Meldeaktezl

2022-09-06

Write an equation of the line that passes through the point (2,-1) and is perpendicular to the line $$2x-3y=5$$

Caiden Brewer

Beginner2022-09-07Added 5 answers

Given that,

Point (2,-1)

Perpendicular line equation is 2x-3y=5

$$\Rightarrow 3y=2x-5\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{2}{3}x-\frac{5}{3}$$

so, slope $$={m}_{1}=\frac{2}{3}$$

Two lines are prependicilar then

$${m}_{1}\times {m}_{2}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3}\times {m}_{1}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}=-\frac{3}{2}$$

The equation of line passes through $$({x}_{1},{y}_{1})$$ and slope m is $$y-{y}_{1}=m(x-{x}_{1})$$

Here, the equation of line passes through (2,-1) and $$m={m}_{2}=-\frac{3}{2}$$ is

$$y-(-1)=-\frac{3}{2}(x-2)\phantom{\rule{0ex}{0ex}}\Rightarrow y+1=-\frac{3}{2}x+\frac{6}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{3}{2}x+3-1\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{3}{2}x+2$$

Point (2,-1)

Perpendicular line equation is 2x-3y=5

$$\Rightarrow 3y=2x-5\phantom{\rule{0ex}{0ex}}\Rightarrow y=\frac{2}{3}x-\frac{5}{3}$$

so, slope $$={m}_{1}=\frac{2}{3}$$

Two lines are prependicilar then

$${m}_{1}\times {m}_{2}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3}\times {m}_{1}=-1\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}=-\frac{3}{2}$$

The equation of line passes through $$({x}_{1},{y}_{1})$$ and slope m is $$y-{y}_{1}=m(x-{x}_{1})$$

Here, the equation of line passes through (2,-1) and $$m={m}_{2}=-\frac{3}{2}$$ is

$$y-(-1)=-\frac{3}{2}(x-2)\phantom{\rule{0ex}{0ex}}\Rightarrow y+1=-\frac{3}{2}x+\frac{6}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{3}{2}x+3-1\phantom{\rule{0ex}{0ex}}\Rightarrow y=-\frac{3}{2}x+2$$

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