The equation of the line passing through the point of intersection of the lines $2 x+3 y+6=0,3 x-y-13=0$ and…
The equation of the line passing through the point of intersection of the lines $2 x+3 y+6=0,3 x-y-13=0$ and parallel to the line $3 x-4 y+5=0$ is
- $3 x-4 y+75=0$
- $3 x-4 y+15=0$
- $3 x-4 y+25=0$
- $3 x-4 y-25=0$
Solution
Let $L_1 \equiv 2 x+3 y+6=0$
$
L_2 \equiv 3 x-y-13=0
$
Equation of line passing through the point of intersection.
$
\begin{array}{r}
L_1+\lambda L_2=0 \\
(2 x+3 y+6)+\lambda(3 x-y-13)=0 \\
(2+3 \lambda) x+(3-\lambda) y+6-13 \lambda=0
\end{array}
$
Which is parallel to the line
$
3 x-4 y+5=0
$
On comparing the coefficient, we get
$
\begin{aligned}
\frac{2+3 \lambda}{3}=\frac{3-\lambda}{-4} & =\frac{6-13 \lambda}{5} \\
-8-12 \lambda & =9-3 \lambda
\end{aligned}
$
$\lambda=-\frac{17}{9}$ putting this value in required equation, we get
$
\begin{array}{rlrl}
\left(2-\frac{51}{9}\right) x+\left(3+\frac{17}{9}\right) y+6+\frac{221}{9} & =0 \\
\Rightarrow & & -33 x+44 y+275 & =0 \\
\Rightarrow & & 3 x-4 y-25 & =0 .
\end{array}
$
Asked in: AP EAMCET 2018 (24 Apr Shift 1)
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