Question

What is the equation of the line that passes through the point (4,-5) and is perpendicular to 2x-5y= -10?

Answer 1

`y=-5/2x+5`

Explanation:

Rewrite the equation of the line we must be perpendicular to as `y=(2x+10)/5= 2/5 x + 2`. This is the slope-intercept form, and indeed we can see that the slope is `m=2/5`, and the intercept is `q=2` (even if we don't care about it in this specific case).

A line with slope `n` is perpendicular to a line with slope `m` if and only if the following equation holds:

`n=-1/m`.

In our case, the slope must be `-1/(2/5)=-5/2`.

So, now we know everything we need, since the slope and a known point identify a line uniquely: we can find the equation with the formula

`y-y_0=m(x-x_0)`, if `m` is the slope of the line and `(x_0,y_0)` is the known point. Plugging the values, we have

`y+5=-5/2(x-4)`, which we can adjust into

`y=-5/2x+5`