Question

Write an equation that is perpendicular to 2x -5y =5 and passes through the point (2,-9).

Answer 1

Consider that the equation of a line with slope 'm' and y-intercept 'c' is given by,

`y=mx+c`

Convert the given equation,

`\begin{gathered} 2x-5y=5 \\ 5y=2x-5 \\ y=(2)/(5)x-1 \end{gathered}`

So the slope of the given equation is,

`m=(2)/(5)`

Let the equation of the perpendicular line be,

`y=m^(\prime)x+c^(\prime)`

For the lines to be parallel, the product of slopes must be -1.

`\begin{gathered} m^(\prime)* m=-1 \\ m^(\prime)*(2)/(5)=-1 \\ m^(\prime)=(-5)/(2) \end{gathered}`

So the equation becomes,

`y=(-5)/(2)x+c^(\prime)`

Given that the perpendicular passes through the point (2,-9),

`\begin{gathered} -9=(-5)/(2)(2)+c^(\prime) \\ -9=-5+c^(\prime) \\ c^(\prime)=-4 \end{gathered}`

Substitute the value in the equation,

`\begin{gathered} y=(-5)/(2)x-4 \\ 2y=-5x-8 \\ 5x+2y+8=0 \end{gathered}`

Thus, the equation of the perpendicular line is 5x + 2y + 8 = 0 .