Question

What is the equation for a line passing through (-2,5) perpendicular to y - 3x = 8

Answer 1

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

`y=mx+c`

Consider the given equation of line,

`\begin{gathered} y-3x=8 \\ y=3x+8 \end{gathered}`

Comparing the coefficient, it is found that the slope of the given line is 3,

`m=3`

Let 's' be the slope of the line which is perpendicular to this line.

Consider that two lines will be perpendicular if their product of slopes is -1,

`\begin{gathered} m* s=-1 \\ 3* s=-1 \\ s=(-1)/(3) \end{gathered}`

So the slope of the perpendicular line is given by,

`y=(-1)/(3)x+c`

Now, it is given that this line passes through the point (-2,5), so it must satisfy the equation of the line,

`\begin{gathered} 5=(-1)/(3)(-2)+c_{} \\ 5=(2)/(3)+c \\ c=5-(2)/(3) \\ c=(13)/(3) \end{gathered}`

Substitute the value of 'c' to get the final equation,

`\begin{gathered} y=(-1)/(3)x+(13)/(3) \\ 3y=-x+13 \\ x+3y=13 \end{gathered}`

Thus, the required equation of the perpendicular line is x + 3y = 13 .