Answer:
[tex]\large\boxed{\sf 3x - 5y +10=0}[/tex]
Step-by-step explanation:
A equation is given to us and we need to find out the equation of the line which is perpendicular to the given line and passes through (-20,-10). The given line is ,
[tex]\longrightarrow 5x + 3y = 30 \\[/tex]
Convert this into slope intercept form of the line , which is y = mx + c.
[tex]\longrightarrow 3y = -5x +30 \\[/tex]
Divide both sides by 3,
[tex]\longrightarrow y =\dfrac{-5}{3}x+\dfrac{30}{3}\\[/tex]
Simplify ,
[tex]\longrightarrow y=-\dfrac{5}{3}x + 10 \\[/tex]
On comparing it to slope intercept form, we have ;
[tex]\longrightarrow m= \dfrac{-5}{3} \\[/tex]
Now as we know that the product of slopes of two perpendicular lines is -1 . Therefore the slope of the perpendicular line will be negative reciprocal of the slope of first line. As ,
[tex]\longrightarrow m_{\perp}=-\bigg(\dfrac{-3}{5}\bigg) = \dfrac{3}{5} \\[/tex]
Now we may use the point slope form of the line which is ,
[tex]\longrightarrow y - y_1 = m(x-x_1) \\[/tex]
Substitute the respective values ,
[tex]\longrightarrow y -(-10) = \dfrac{3}{5}\{ x -(-20)\}\\[/tex]
Simplify the brackets ,
[tex]\longrightarrow y +10 =\dfrac{3}{5}(x+20) \\[/tex]
Cross multiply ,
[tex]\longrightarrow5( y +10)= 3(x+20)\\[/tex]
Distribute ,
[tex]\longrightarrow 5y +50 = 3x +60\\[/tex]
Subtract (5y +50) to both sides ,
[tex]\longrightarrow 3x + 60 -5y -50=0 \\[/tex]
Simplify ,
[tex]\longrightarrow \underline{\underline{ 3x -5y +10=0}} \\[/tex]
