Question

If the perpendicular drawn from the point (2,-3) to the straight line $4x-3y+8=0$ meets it at M(a,b) and $a^3-b^3=k^3$, then $k =$

• A. 1
• B. -1
• C. 2
• D. -2

Hint:

Memorizing the formula for the foot of the perpendicular can save a lot of time compared to the method of finding the equation of the perpendicular line and then solving the system of two linear equations.

Answer 1

The Correct Option is D

The point M(a,b) is the foot of the perpendicular from P(2,-3) to the line $L: 4x-3y+8=0$.
We can find the coordinates of M(a,b) using the formula for the foot of the perpendicular from $(x_1, y_1)$ to the line $Ax+By+C=0$:
$\frac{a-x_1}{A} = \frac{b-y_1}{B} = -\frac{Ax_1+By_1+C}{A^2+B^2}$.
Here, $(x_1, y_1)=(2,-3)$ and the line is $4x-3y+8=0$, so $A=4, B=-3, C=8$.
$\frac{a-2}{4} = \frac{b-(-3)}{-3} = -\frac{4(2)-3(-3)+8}{4^2+(-3)^2}$.
$\frac{a-2}{4} = \frac{b+3}{-3} = -\frac{8+9+8}{16+9} = -\frac{25}{25} = -1$.
Now we can find $a$ and $b$ separately.
$\frac{a-2}{4} = -1 \implies a-2 = -4 \implies a = -2$.
$\frac{b+3}{-3} = -1 \implies b+3 = 3 \implies b = 0$.
So, the coordinates of M are $(a,b) = (-2, 0)$.
We are given the relation $a^3 - b^3 = k^3$.
Substitute the values of $a$ and $b$:
$(-2)^3 - (0)^3 = k^3$.
$-8 - 0 = k^3$.
$k^3 = -8$.
$k = -2$.