Mathematics - Straight line Question with Solution | TestHub

Given equation of lines are $\left(a^3+3\right)x+ay+a-3=0$ and $\left(a^5+2\right)x+(a+2)y+2a+3=0$ (a real) Since point of intersection of lines lies on $\mathrm{y}$-axis. $\therefore$ Put $x=0$ in each equation, we get $a\mathrm{y}+a-3=0$ and $(a+2)y+2a+3=0$ On solving these we get $$\begin{array}{rl}& (a+2)(a-3)-a(2a+3)=0\\ & \Rightarrow a^2-a-6-2a^2-3a=0\\ & \Rightarrow -a^2-4a-6=0\Rightarrow a^2+4a+6=0\\ & \Rightarrow a=\frac{-4\pm \sqrt{16-24}}{2}=\frac{-4\pm \sqrt{-8}}{2}\\ & (\text{ not real) }\end{array}$$ This shows that the point of intersection of the lines lies on the $y$-axis for no value of ' $a$ '.