Question

If the line (2x + 3y + 4) +λ (6x – y + 12) = 0 is perpendicular to the line 7x + y – 4 = 0, then λ = _____________.

Answer 1

For given line,

$$\begin{gathered} \left(2x+3y+4\right)+\lambda \left(6x-y+12\right)=0 \\ \mathrm{i}.\mathrm{e}.\left(2+6\lambda \right)x+\left(3-\lambda \right)y+4+12\lambda =0 \\ \mathrm{i}.\mathrm{e}.\left(3-\lambda \right)y=-\left(2+6\lambda \right)x-4-12\lambda \\ \mathrm{i}.\mathrm{e}.y=\frac{-2\left(1+3\lambda \right)}{\left(3-\lambda \right)}x+\frac{\left(-4\right)\left(1+3\lambda \right)}{3-\lambda } \\ \mathrm{i}.\mathrm{e}.\mathrm{slope}\mathrm{is}\frac{-2\left(1+3\lambda \right)}{3-\lambda } \\ \mathrm{which}\mathrm{is}\mathrm{perpendicular}\mathrm{to}7x+y-4=0 \\ \mathrm{i}.\mathrm{e}.y=-7x+4 \\ \therefore \frac{\left(-2\right)\left(1+3\lambda \right)}{3-\lambda }\times \left(-7\right)=-1 \\ 14\left(1+3\lambda \right)=-3+\lambda \\ 14+42\lambda =-3+\lambda \\ \mathrm{i}.\mathrm{e}.41\mathrm{\lambda }=-17 \\ \mathrm{i}.\mathrm{e}.\lambda =\frac{-17}{41} \end{gathered}$$