Question

If $u=a_{1}x+b_{1}y+c_1=0,$ $v=a_{2}x+b_{2}y+c_2=0$ and
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ , then the curve
$u+kv=0$ is

• A. The same straight line 𝑢
• B. Not a straight line
• C. Different straight line
• D. None of these

Answer 1

The correct option is A The same straight line 𝑢

Given: $u=a_{1}x+b_{1}y+c_1=0,v=a_{2}x+b_{2}y+c_2=0$
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=m\left(say\right)$
$a_1=m{a}_2,b_1=m{b}_2,c_1=m{c}_2$
Now,
𝑢+𝑘𝑣=0
$\therefore a_{1}x+b_{1}y+c_1+k(a_{2}x+b_{2}y+c_2)=0$
$\therefore x(a_1+k{a}_2)+y(b_1+k{b}_2)+c_1+k{c}_2=0$
$\therefore (m+k)[a_{2}x+b_{2}y+c_2]=0$
$\therefore a_{2}x+b_{2}y+c_2=0$ $[\because m+k\ne 0]$
$\therefore \frac{1}{m}(a_{1}x+b_{1}y+c_1)=0$
$\therefore a_{1}x+b_{1}y+c_1=0$
Hence, the curve $u+kv=0$ is same as straight line 𝑢.