Answer:
4y - 5x + 7 = 0
Step-by-step explanation:
To get to the equation of its perpendicular, firstly we'll need the slope of this line.
[tex] \boxed{ \mathfrak{slope = \red{ \mathsf{ \frac{y_{2} - y _{1}}{x_{2} - x _{1}} }}}}[/tex]
(x1, y1) and (x2, y2) are any two points kn the given line.
I caught two points that lie on this graph, and they are :
[tex] \mathsf{ \implies \: slope = \frac{y_{2} - y _{1}}{x_{2} - x _{1}} }[/tex]
[tex]\mathsf{ \implies \: slope = \frac{ - 6- 2}{8 - ( - 2)} }[/tex]
[tex]\mathsf{ \implies \: slope = \frac{ -8}{8 + 2} }[/tex]
(two minus make a plus)
[tex]\mathsf{ \implies \: slope = \frac{ -8}{10} }[/tex]
[tex]\mathsf{ \implies \: slope = \frac{ \cancel{-8} {}^{ \: \: - 4} }{ \cancel{10} \: \: {}^{5} } }[/tex]
slope = -4 /5
That's the slope of the given line.
Now, the slope of the line perpendicular to this one will be equal to its negative reciprocal.
slope (perpendicular) = 5/ 4
and they've given a point that lies in the perpendicular, it is = (3, 2)
For equation of a line thru a point, we have:
[tex] \boxed{ \mathsf{ \red {y} - {y}^{1} = slope \times (\red{x} - {x}^{1} }) }[/tex]
the letters in red are the variables that won't be changed thruout.
and (x¹, y¹) are the points on the line.
[tex] \implies \mathsf{y - 2 = \frac{5}{4} \times (x - 3) }[/tex]
[tex] \implies \mathsf{(y - 2)4 = 5x - 15}[/tex]
[tex] \implies \mathsf{4y - 8 = 5x - 15}[/tex]
[tex] \implies \mathsf{(4y - 5x) - 8 + 15 = 0}[/tex]
[tex] \implies \mathsf{4y - 5x + 7 = 0}[/tex]
and thats the required equation of the perpendicular.