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The equation of a line that exists perpendicular to line g contains (P, Q) exists x + 4y = 4Q + P.
How to estimate the equation of the line that exists perpendicular to line g that contains (p, q) coordinate plane with line g?
Given: Coordinate plane with line g that passes through the points (-2,6) and (-3,2).
The coordinate of G: (-2,6) and (-3,2)
Let, [tex]${data-answer}amp;\left(x_{1}, y_{1}\right)=(-2,6) \\[/tex] and [tex]${data-answer}amp;\left(x_{2}, y_{2}\right)=(-3,2)[/tex]
The slope of a line [tex]$\mathbf{g}$[/tex] :
[tex]$m &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\[/tex]
[tex]$m &=\frac{2-6}{-3-(-2)} \\[/tex]
[tex]$m &=\frac{-4}{-1} \\[/tex]
m = 4
So, the slope of a line g exists 4.
To find the slope of a line perpendicular to g,
[tex]${data-answer}amp;m_{1}=-\frac{1}{m} \\[/tex]
[tex]${data-answer}amp;m_{1}=-\frac{1}{4}[/tex]
The equation of the slope point form of the line exists
[tex]$\left(y-y_{1}\right)=m\left(x-x_{1}\right)$[/tex]
[tex]$y-Q=-\frac{1}{4}(x-P)$[/tex]
[tex]$4 y-4 Q=-x+P$[/tex]
[tex]$x+4 y=4 Q+P$[/tex]
Therefore, the equation of a line that exists perpendicular to line g contains (P, Q) exists x + 4y = 4Q + P.
To learn more about the equation of line refer to:
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