Question

Find the equation of the line that is perpendicular to 5x+3y=30 and passes through (-20, -10).

Answer 1

`\large\boxed{\sf 3x - 5y +10=0}`

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 ,

`\longrightarrow 5x + 3y = 30 \\`

Convert this into slope intercept form of the line , which is y = mx + c.

`\longrightarrow 3y = -5x +30 \\`

Divide both sides by 3,

`\longrightarrow y =(-5)/(3)x+(30)/(3)\\`

Simplify ,

`\longrightarrow y=-(5)/(3)x + 10 \\`

On comparing it to slope intercept form, we have ;

`\longrightarrow m= (-5)/(3) \\`

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 ,

`\longrightarrow m_(\perp)=-\bigg((-3)/(5)\bigg) = (3)/(5) \\`

Now we may use the point slope form of the line which is ,

`\longrightarrow y - y_1 = m(x-x_1) \\`

Substitute the respective values ,

`\longrightarrow y -(-10) = (3)/(5)\{ x -(-20)\}\\`

Simplify the brackets ,

`\longrightarrow y +10 =(3)/(5)(x+20) \\`

Cross multiply ,

`\longrightarrow5( y +10)= 3(x+20)\\`

Distribute ,

`\longrightarrow 5y +50 = 3x +60\\`

Subtract (5y +50) to both sides ,

`\longrightarrow 3x + 60 -5y -50=0 \\`

Simplify ,

`\longrightarrow \underline{\underline{ 3x -5y +10=0}} \\`