Question 116

If the lines $$3x - ky + 4 = 0$$ and $$4x + y + 2 = 0$$ are perpendicular to each other, then $$k^2 - 12k + 4 =$$

Solution

Slope of the line  $$ax+by+c=0$$ is  $$\frac{-a}{b}$$

Slope of the line $$3x - ky + 4 = 0$$ is $$\frac{-3}{-k}=\frac{3}{k}$$

Slope of the line $$4x + y + 2 = 0$$ is $$\frac{-4}{1}=-4$$

The product of slope of two lines perpendicular to each other is -1.

$$\Rightarrow$$  $$\frac{3}{k}\times\left(-4\right)=-1$$

$$\Rightarrow$$  $$k=12$$

$$\therefore\ $$ $$k^2-12k+4=12^2-12\left(12\right)+4=4$$

Hence, the correct answer is Option B


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