Question:hard

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

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Standard parametric coordinates $(x_1 + r\cos\theta, y_1 + r\sin\theta)$ make distance-along-direction problems incredibly simple to solve.
Updated On: Jun 3, 2026
  • $3\sqrt{2}$
  • $2\sqrt{2}$
  • $\sqrt{2}$
  • $5\sqrt{2}$
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The Correct Option is A

Solution and Explanation

Step 1: Understand the special distance.
We do not want the usual perpendicular distance. We want the distance from $(1, 2)$ to the line, but measured along a particular direction parallel to $x - y = 0$. So we travel in that direction until we hit the line.

Step 2: Find the direction angle.
The guide line $x - y = 0$ has slope $1$, which means it makes a $45^\circ$ angle. So $\cos 45^\circ = \frac{1}{\sqrt{2}}$ and $\sin 45^\circ = \frac{1}{\sqrt{2}}$.

Step 3: Write a moving point.
Starting at $(1, 2)$ and moving a distance $r$ in that direction gives the point.
\[ \left(1 + \frac{r}{\sqrt{2}},\ 2 + \frac{r}{\sqrt{2}}\right) \]

Step 4: Make it lie on the target line.
This point must satisfy $3x + 4y - 32 = 0$. Substitute.
\[ 3\left(1 + \frac{r}{\sqrt{2}}\right) + 4\left(2 + \frac{r}{\sqrt{2}}\right) - 32 = 0 \]

Step 5: Simplify the equation.
Expand: $3 + \frac{3r}{\sqrt{2}} + 8 + \frac{4r}{\sqrt{2}} - 32 = 0$. Combine constants and the $r$ terms.
\[ \frac{7r}{\sqrt{2}} - 21 = 0 \]

Step 6: Solve for $r$.
Move and multiply.
\[ \frac{7r}{\sqrt{2}} = 21 \implies r = \frac{21\sqrt{2}}{7} = 3\sqrt{2} \]
\[ \boxed{3\sqrt{2}} \]
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