Question

Find the equation of a line perpendicular to the line $x-2y+3=0$ and passing through the point $(1,–2)$.

Answer 1

Step-$1$: Slope of given lines

Let $AB$ be the given line, whose equation
is $x-2y+3=0$

Let the given point be $P$ where $P=(1,-2$)

Let line $CD$ be perpendicular to line $AB$ and passing through point $P(1,-2)$

Lets first calculate slope of line $AB$

$\Rightarrow x-2y+3=0$

$\Rightarrow 2y=x+3$

$\Rightarrow y=\frac{x+3}{2}$

$\Rightarrow y=\frac{x}{2}+\frac{3}{2}$

$\Rightarrow y=\frac{1}{2}x+\frac{3}{2}$

The above equation is of the form $y=mx+c$

Thus, slope of line $AB=\frac{1}{2}$

Step-$2$: Required equation of line

We know that product of slope of perpendicular lines is $-1$

Here, line $AB$ is perpendicular to line $CD$

$\therefore$ Slope of line $AB\times$ Slope of line $CD=-1$

$\Rightarrow \frac{1}{2}\times$ Slope of line $CD=-1$

$\Rightarrow$ Slope of line $CD=-1\times \frac{2}{1}$

$\Rightarrow$ Slope of line $CD=-2$

Thus, line $CD$ has a slope $-2$, passes through point $P(1,-2)$

We know that equation of line having slope $m$ and passing through point $(x_1,y_1)$ is

$y-y_1=m(x-x_1)$

Putting values for line $CD$ $x_1=1,y_1=-2andm=-2$

$\Rightarrow y-(-2)=-2(x-1)$

$\Rightarrow y+2=-2(x-1)$

$\Rightarrow y+2=-2x+2$

$\Rightarrow y+2+2x-2=0$

$\Rightarrow y+2x=0$

$\Rightarrow y=-2x$

Final Answer:

Therefore, the required equation of line is $y=-2x$.