Question

The distance of point \( (1, 2) \) from line \( x + y + 2 = 0 \) measured along the line parallel to \( 2x - y = 5 \) is equal to:

• A. \( \frac{125}{3} \)
• B. \( \frac{125}{\sqrt{3}} \)
• C. \( \frac{5\sqrt{5}}{3} \)
• D. \( \frac{5\sqrt{3}}{3} \)

Hint:

When measuring distance along a line not perpendicular, use projection of the perpendicular vector along the desired direction.

Answer 1

The Correct Option is C

We need distance from point \( P(1, 2) \) to line \( L: x + y + 2 = 0 \), but measured along a direction parallel to line \( 2x - y = 5 \).
Direction vector of this line: \( \vec{d} = (1, 2) \Rightarrow \) direction ratio \( \vec{v} = (2, 1) \)
Normal to the line \( x + y + 2 = 0 \): \( \vec{n} = (1, 1) \)
Step 1: Projection of vector \( \vec{AP} \) onto \( \vec{v} \), where \( A \) is the foot of perpendicular from P onto the line. Use formula:
\[ \text{Distance} = \left| \frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}} \right| \cdot \sec \theta \]
Where \( \theta \) is the angle between \( \vec{n} \) and direction of projection.
Angle \( \theta \) between \( \vec{n} = (1,1) \) and \( \vec{v} = (2,1) \):
\[ \cos \theta = \frac{1 \cdot 2 + 1 \cdot 1}{\sqrt{1^2 + 1^2} \cdot \sqrt{2^2 + 1^2}} = \frac{3}{\sqrt{2} \cdot \sqrt{5}} = \frac{3}{\sqrt{10}} \Rightarrow \sec \theta = \frac{\sqrt{10}}{3} \]
Now, perpendicular distance:
\[ \left| \frac{1 + 2 + 2}{\sqrt{1^2 + 1^2}} \right| = \frac{5}{\sqrt{2}} \Rightarrow \text{Projected distance} = \frac{5}{\sqrt{2}} \cdot \frac{\sqrt{10}}{3} = \frac{5 \sqrt{5}}{3} \]