Line Distance Problem - Detailed Solution and Interactive Quiz

Line Distance Problem [IIT 1965]

Question: If the slope of a line passing through the point \( A(3,\,2) \) is \( \frac{3}{4} \), then the points on the line which are 5 units away from \( A \) are:

(a) \( (5,\,5) \), \( (-1,\,-1) \)

(b) \( (7,\,5) \), \( (-1,\,-1) \)

(c) \( (5,\,7) \), \( (-1,\,-1) \)

(d) \( (7,\,5) \), \( (1,\,1) \)

Detailed Step-by-Step Explanation

Step 1: Write the equation of the line passing through \( A(3,2) \) with slope \( \frac{3}{4} \). Using the point-slope form:

\( y - 2 = \frac{3}{4}(x - 3) \)

Simplify the equation:
\( y = \frac{3}{4}x - \frac{9}{4} + 2 = \frac{3}{4}x - \frac{9}{4} + \frac{8}{4} = \frac{3}{4}x - \frac{1}{4} \)

Step 2: Let \( P(x, y) \) be a point on the line. Then \( y = \frac{3}{4}x - \frac{1}{4} \). The distance between \( A(3,2) \) and \( P(x,y) \) is given by:

\( \sqrt{(x-3)^2 + \left(y-2\right)^2} = 5 \)

Substitute \( y \) from the line equation:

\( \sqrt{(x-3)^2 + \left(\frac{3}{4}x - \frac{1}{4} - 2\right)^2} = 5 \)

Simplify \( \frac{3}{4}x - \frac{1}{4} - 2 = \frac{3}{4}x - \frac{1}{4} - \frac{8}{4} = \frac{3}{4}x - \frac{9}{4} \).

So, the equation becomes:

\( \sqrt{(x-3)^2 + \left(\frac{3}{4}x - \frac{9}{4}\right)^2} = 5 \)

Step 3: Square both sides to eliminate the square root:

\( (x-3)^2 + \left(\frac{3}{4}x - \frac{9}{4}\right)^2 = 25 \)

Notice that \( \frac{3}{4}x - \frac{9}{4} = \frac{3}{4}(x-3) \). Thus:

\( (x-3)^2 + \left(\frac{3}{4}(x-3)\right)^2 = 25 \)

Step 4: Simplify the equation:

\( (x-3)^2 + \frac{9}{16}(x-3)^2 = \left(1 + \frac{9}{16}\right)(x-3)^2 = \frac{25}{16}(x-3)^2 = 25 \)

Solve for \( (x-3)^2 \):

\( (x-3)^2 = 25 \times \frac{16}{25} = 16 \)

Taking the square root:

\( x - 3 = \pm 4 \)

Therefore, the possible \( x \)-coordinates are:
\( x = 3 + 4 = 7 \) or \( x = 3 - 4 = -1 \)

Step 5: Find the corresponding \( y \)-coordinates:
For \( x = 7 \):

\( y = \frac{3}{4}(7) - \frac{1}{4} = \frac{21}{4} - \frac{1}{4} = \frac{20}{4} = 5 \)

For \( x = -1 \):

\( y = \frac{3}{4}(-1) - \frac{1}{4} = -\frac{3}{4} - \frac{1}{4} = -\frac{4}{4} = -1 \)

Thus, the points on the line that are 5 units away from \( A(3,2) \) are:
\( (7,\,5) \) and \( (-1,\,-1) \)

This corresponds to option (b).