Question

The distance of the point \((1,1)\) from the line \(2x - 3y - 4 = 0\) in the direction of the line \(x + y = 1\), is

• A. \(\sqrt{2}\)
• B. \(\frac{5\sqrt{2}}{2}\)
• C. \(\frac{1}{\sqrt{2}}\)
• D. \(\frac{1}{2}\)

Hint:

The distance along a direction is not the perpendicular distance; use parametric form.

Answer 1

The Correct Option is A

To find the distance of the point \((1, 1)\) from the line \(2x - 3y - 4 = 0\) in the direction of the line \(x + y = 1\), we proceed as follows:

1. First, we need the point on the line \(2x - 3y - 4 = 0\) that is closest to the point \((1, 1)\). To do that, we take a point on the line as \((x_1, y_1)\).
2. Since the distance from the point \((1, 1)\) to the line should be perpendicular to the line, the direction of this distance vector (a, b) should be perpendicular to the direction vector of the line \(2x - 3y - 4 = 0\).
3. The normal vector to the line \(2x - 3y - 4 = 0\) is \((2, -3)\). We take a point on the line, say \((2, 0)\), to find the perpendicular distance direction vector.
4. Now, calculate the dot product of the directional vector of distance and direction of the given line \(x + y = 1\). The direction vector is perpendicular to the normal \((1, -1)\). Normalize it: \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\) (by dividing by magnitude).
5. The shortest distance from point \((1, 1)\) to the line ax + by + c = 0 is given by: \(\frac{|2(1) - 3(1) - 4|}{\sqrt{2^2 + (-3)^2}}\).
   • Calculate \(|2(1) - 3(1) - 4| = |2 - 3 - 4| = |-5| = 5\).
   • Calculate the magnitude \(\sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\).
   • The shortest distance is \(\frac{5}{\sqrt{13}}\).
6. Travel this distance in the direction of the line \(x + y = 1\) direction, i.e., along vector \(\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\).
   • The projected distance in this direction is \(5 \cdot \frac{1}{\sqrt{13}} \cdot \sqrt{2}\).
   • This simplifies to \(\sqrt{2}\).
7. Thus, the required distance of the point \((1, 1)\) from the line \(2x - 3y - 4 = 0\) in the direction of \(x + y = 1\) is \(\sqrt{2}\).