Question

The equation of the parallel to the line $2x-3y=1$ and passing through the point of the segment joining the points $(1,3)$ and $(1,-7)$, is

• A. $2x-3y+8=0$
• B. $2x-3y=8$
• C. $2x-3y+4=0$
• D. $2x-3y=4$

Answer 1

The correct option is A $2x-3y+8=0$

Given that,

$2x-3y=1$--------(i)

$\Rightarrow -3y-1-2x$

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

$\left[y=mx+c\right]$

$\therefore m=\frac{2}{3}$

equation of line $AB$

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

$\frac{y-3}{x-1}=\frac{-7-3}{1-1}$

$\frac{y-3}{x-1}=0$

$\Rightarrow y-3=0$

$\Rightarrow y=3$

From (i)

$2x-3y=1$

$2x-3\times 3=1$

$\Rightarrow 2x=10$

$\Rightarrow x=5$

$p=(x.y)=(5,3)$

equation of $MN$ at $(5,3)$

$y-y_1=m\left(x-x_1\right)$

$y-3=\frac{2}{3}\left(x-5\right)$

$\Rightarrow 3\left(y-3\right)=2\left(x-5\right)$

$\Rightarrow 3y-9=2x-10$

$\Rightarrow 3y-2x=-10+9$

$\Rightarrow 3y-2x=-1$

$2x-3y-1=0$

Which is required equation of line.