Question

$m\equiv a_{1}x+b_{1}y+c_1=0$ and $l\equiv a_{2}x+b_{2}y+c_2=0$ are two straight lines such that $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ then $m+kl=0,k\in R$ is :

• A. a straight line different from $m$ and $l$
• B. not a straight line
• C. is a straight line concurrent with $m$ and $l$
• D. the same straight line $m=0$

Answer 1

The correct option is D the same straight line $m=0$

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

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

$l\&m$ are two straight lines such that $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\lambda$

So, $l\equiv \frac{a_{1}x+b_{1}y+c_1}{\lambda }=0$

$\therefore m+kl=x(a_1+k\frac{a_1}{\lambda })+y(b_1+k\frac{b_1}{\lambda })+c_1+k\frac{c_1}{\lambda }=0$

$=\frac{(\lambda +k)}{\lambda }{a}_{1}x+\frac{(\lambda +k)}{\lambda }{b}_{1}y+\frac{(\lambda +k)}{\lambda }{c}_1=0$

$\frac{(\lambda +k)}{\lambda }(a_{1}x+b_{1}y+c_1)=0$

The same straight line $m=0.$