Question

The nearest point on the line $x + y - 3 = 0$ from the point $(3, -2)$ is:

• A. $(3, 5)$
• B. $(4, 1)$
• C. $(3, -5)$
• D. $(4, -1)$
• E. $(5, -1)$

Hint:

Since the nearest point MUST lie on the line, eliminate options that don't satisfy $x + y - 3 = 0$. Here, $4 + (-1) - 3 = 0$ works, while $(3, 5)$ or $(3, -5)$ do not. This often allows you to find the answer without the formula.

Answer 1

The Correct Option is D

Concept: The "nearest point" on a line from an external point is the foot of the perpendicular dropped from that point to the line. For a point $(x_1, y_1)$ and line $ax + by + c = 0$, the foot of the perpendicular $(h, k)$ is given by: \[ \frac{h - x_1}{a} = \frac{k - y_1}{b} = -\frac{(ax_1 + by_1 + c)}{a^2 + b^2} \]

Step 1: Identify the coefficients and substitute.
Point $(x_1, y_1) = (3, -2)$. Line $x + y - 3 = 0$ ($a=1, b=1, c=-3$). Calculate the ratio: \[ R = -\frac{(1(3) + 1(-2) - 3)}{1^2 + 1^2} = -\frac{(3 - 2 - 3)}{2} = -\frac{-2}{2} = 1 \]

Step 2: Solve for $h$ and $k$.
Using $\frac{h - 3}{1} = 1
\Rightarrow h - 3 = 1
\Rightarrow h = 4$. Using $\frac{k - (-2)}{1} = 1
\Rightarrow k + 2 = 1
\Rightarrow k = -1$.

Step 3: Conclusion.
The nearest point is $(4, -1)$.