# How do you find the equation of the line t hat is perpendicular to x - 2y = -10 and passes through the point ( -1 , -3 )?

Sep 12, 2016

$2 x + y + 5 + 0$.

#### Explanation:

Consider the following Results :

$\text{The eqn. of a line L' "bot" L : "ax+by+c=0" is given by}$

$L ' : b x - a y + k = 0 , w h e r e , {a}^{2} + {b}^{2} \ne 0$.

$\text{In addition, L' passes through a pt. "(x_1,y_1)," then,}$

$L ' : b \left(x - {x}_{1}\right) - a \left(y - {y}_{1}\right) = 0$.

If we use these Results, we can immediately write the eqn. of reqd.

line as$2 \left(x + 1\right) + \left(y + 3\right) = 0 , i . e . , 2 x + y + 5 + 0$.

But, for the help of these Results, we can derive the eqn. as under :

Eqn. of the given line $: x - 2 y = - 10 , i . e . , 2 y = x + 10 , \mathmr{and} , y = \frac{1}{2} x + 5$.

So, the slope of this line is $\frac{1}{2}$, & hence, the slope reqd line (perp.

to given line) must be $- 2$. We also have a pt.(-1,-3) on this line.

Hence, by the Slope-Pt. Form of line, the eqn. of reqd. line is,

$y + 3 = - 2 \left(x + 1\right)$, which is the same as obtained earlier!.

Enjoy Maths.!