Question

The distance of the point $(1, 2)$ from the line $3x + 4y - 32 = 0$ measured parallel to the line $x - y = 0$ is:

• A. $3\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\sqrt{2}$
• D. $5\sqrt{2}$

Hint:

Standard parametric coordinates $(x_1 + r\cos\theta, y_1 + r\sin\theta)$ make distance-along-direction problems incredibly simple to solve.

Answer 1

The Correct Option is A

Step 1: Concept
The coordinates of any point $Q$ at a distance $r$ from $P(x_1, y_1)$ along a line making an angle $\theta$ with the positive direction of the x-axis are $(x_1 + r\cos\theta, y_1 + r\sin\theta)$.

Step 2: Meaning
The line is measured parallel to $x - y = 0$, which has a slope $m = 1 \implies \theta = 45^\circ$. Thus, $\cos\theta = \frac{1}{\sqrt{2}}$ and $\sin\theta = \frac{1}{\sqrt{2}}$.

Step 3: Analysis
Let the coordinates of $Q$ be: \[ Q\left(1 + \frac{r}{\sqrt{2}}, \, 2 + \frac{r}{\sqrt{2}}\right) \] Since $Q$ lies on the line $3x + 4y - 32 = 0$: \[ 3\left(1 + \frac{r}{\sqrt{2}}\right) + 4\left(2 + \frac{r}{\sqrt{2}}\right) - 32 = 0 \] \[ 3 + \frac{3r}{\sqrt{2}} + 8 + \frac{4r}{\sqrt{2}} - 32 = 0 \] \[ \frac{7r}{\sqrt{2}} - 21 = 0 \implies \frac{7r}{\sqrt{2}} = 21 \implies r = 3\sqrt{2} \]

Step 4: Conclusion
The required distance of the point along the given direction is $3\sqrt{2}$. Final Answer: (A)