# How do you find a standard form equation for the line with (-1,-2) ; perpendicular to the line 2x+5y+8=0?

Mar 31, 2017

$2 y - 5 x - 1 = 0$
or
$5 x - 2 y + 1 = 0$

#### Explanation:

Always keep in mind that when two lines are perpendicular, the product of their slopes is -1.

Proof :-

slope intercept equation of a straight line is given as:-
$y = m x + c$
$\to$ $m =$ slope
$\to$ $c =$ y intercept (y coordinate of the point where the line intersects the y axis) which is a constant for a given line.

given line $2 x + 5 y + 8 = 0$ can be written as
$5 y = - 8 - 2 x$ $\implies$ $y = - \frac{2}{5} x - \frac{8}{5}$
comparing with slope intercept form, slope ${m}_{1} = - \frac{2}{5}$
let slope of second line be ${m}_{2}$.
$\therefore$ ${m}_{1} \cdot {m}_{2} = - 1$ (given lines are perpendicular)
$\implies {m}_{2} = \frac{5}{2}$

$\therefore$ equation of line we have to find is given by:-

$y = \frac{5}{2} x + {c}_{2}$ where

since it passes through $\left(- 1 , - 2\right)$ substituting this point in the equation,
$- 2 = \frac{5}{2} \left(- 1\right) + {c}_{2}$
$\implies {c}_{2} = \frac{5}{2} - 2 = \frac{1}{2}$

hence equation becomes,
$y = \frac{5}{2} x + \frac{1}{2}$
$i . e . 2 y - 5 x - 1 = 0$

Hence required equation is $2 y - 5 x - 1 = 0$ or $5 x - 2 y + 1 = 0$