Question

If $9a^2+16b^2-24ab-25c^2=0$, then the family of straight lines $ax+by+c=0$ is concurrent at the points whose co-ordinates are given by______

Answer 1

$9a^2+16b^2-24ab-25c^2=0\left(\text{Given}\right)$

$\Rightarrow {\left(5c\right)}^2={(3a-4b)}^2$

$\Rightarrow c=\pm \frac{3a-4b}{5}$

Case I:- For $c=\frac{3a-4b}{5}$

$ax+by+c=0$

$ax+by+\frac{3a-4b}{5}=0$

$a(x+\frac{3}{5})+b(y-\frac{4}{5})=0$

$(x+\frac{3}{5})+\frac{b}{a}(y-\frac{4}{5})=0$

$\Rightarrow L_1+\lambda L_2=0$

Concurrent point $=$ Intersection point of $L_1$ and $L_2$, i.e.,

$x+\frac{3}{5}=0y-\frac{4}{5}=0\Rightarrow (\frac{-3}{5},\frac{4}{5})$

Case II:- For $c=-\frac{3a-4b}{5}$

$ax+by+c=0$

$ax+by-\frac{3a-4b}{5}=0$

$a(x-\frac{3}{5})+b(y+\frac{4}{5})=0$

$(x-\frac{3}{5})+\frac{b}{a}(y+\frac{4}{5})=0$

$\Rightarrow L_1^{\prime }+\lambda L_2^{\prime }=0$

Concurrent point $=$ Intersection point of $L_1^{\prime }$ and $L_2^{\prime }$, i.e.,

$x-\frac{3}{5}=0y+\frac{4}{5}=0\Rightarrow (\frac{3}{5},\frac{-4}{5})$

Hence the family of given straight lines is concurrent at the points $(\frac{-3}{5},\frac{4}{5})$ and $(\frac{3}{5},\frac{-4}{5})$.