Question

If $3x+4y+2=0$ and $9x+12y+k=0$ represent coincident lines, then find $k$.

Answer 1

The coincident lines are given as,

$3x+4y+2=0$

And

$9x+12y+k=0$

Comparing the equations with the general equations

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

Then,

$a_1=3,b_1=4,c_1=2,a_2=9,b_2=12,c_2=k$

Since, the lines are coincident, then,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\frac{3}{9}=\frac{4}{12}=\frac{2}{k}$

$\frac{1}{3}=\frac{1}{3}=\frac{2}{k}$

$\frac{1}{3}=\frac{2}{k}$

$k=6$