List of practice Questions

The point $(2, 1)$ is translated parallel to the line $L : x-y = 4$ by $2\sqrt{3}$ units. If the new point $Q$ lies in the third quadrant, then the equation of the line passing through $Q$ and perpendicular to $L$ is :

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2 x - y + z+ 3 = 0$ is the line.