Question

(b) Find the equation of the line which is parallel to 3x - 2y = -4 and passes
through the point of intersection of lines 2x + 3y = 12 and x - y = 1​

Answer 1

Answer:

See below.

Step-by-step explanation:

We are given the two lines,

$\begin{aligned}L_{1}:2x+3y-12=0\\ L_{2}:x-y-1=0\end{aligned}$

The family of lines through the intersection of these lines,

$\begin{aligned}L_{1}+\lambda L_{2}=0\\ 2x+3y-12+\lambda \left( x-y-1\right) =0\\ x\left( 2+\lambda \right) +y\left( 3-\lambda \right) -12-\lambda =0\end{aligned}$

The slope of the above line has to be equal to the slope of the given line. Therefore,

$\begin{aligned}-\dfrac {2+\lambda }{3-\lambda }=\dfrac {3}{2}\\ 4+2\lambda =3\lambda -9\\ \lambda =13\end{aligned}$

The required equation of line is obtained by placing the value of lambda,

$\begin{aligned}x\left( 2+13\right) +y\left( 3-13\right) -12-13=0\\ 15x-10y-25=0\\ 3x-2y-5=0\end{aligned}$

Check that the slope of this line is equal to the given line.