Question

Point \(P(6,4)\) lies on the line \(x - y - 2 = 0\). If \(A(\alpha, \beta)\) and \(B(\gamma, \delta)\) are two points on this line lying on either side of \(P\) at a distance of 4 units from \(P\), then find \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\).

• A. 136
• B. \(\frac{85}{\sqrt{2}}\)
• C. \(23 + \frac{5}{\sqrt{2}}\)
• D. 52

Hint:

Use parametric form of line and distance formula to find points at a given distance.

Answer 1

The Correct Option is A

Step 1: Equation of line
Given line is \[ x - y - 2 = 0 \] Step 2: Points \(A\) and \(B\) lie on the line at distance 4 from \(P\)
Using distance formula and parametric form of line, find \(A\) and \(B\).
If \(P = (6,4)\), direction vector of line is \((1,1)\).
Points \(A\) and \(B\) can be written as \[ A = (6 + t, 4 + t), \quad B = (6 - t, 4 - t) \] Distance \(PA = PB = 4\) gives \[ \sqrt{t^2 + t^2} = 4 \Rightarrow \sqrt{2} t = 4 \Rightarrow t = \frac{4}{\sqrt{2}} = 2 \sqrt{2} \] Step 3: Calculate sum of squares
\[ \alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (6 + 2 \sqrt{2})^2 + (4 + 2 \sqrt{2})^2 + (6 - 2 \sqrt{2})^2 + (4 - 2 \sqrt{2})^2 \] Calculate each term: \[ (6 + 2 \sqrt{2})^2 = 36 + 24 \sqrt{2} + 8 = 44 + 24 \sqrt{2} \] \[ (4 + 2 \sqrt{2})^2 = 16 + 16 \sqrt{2} + 8 = 24 + 16 \sqrt{2} \] \[ (6 - 2 \sqrt{2})^2 = 36 - 24 \sqrt{2} + 8 = 44 - 24 \sqrt{2} \] \[ (4 - 2 \sqrt{2})^2 = 16 - 16 \sqrt{2} + 8 = 24 - 16 \sqrt{2} \] Sum all: \[ (44 + 24 \sqrt{2}) + (24 + 16 \sqrt{2}) + (44 - 24 \sqrt{2}) + (24 - 16 \sqrt{2}) = (44 + 24 + 44 + 24) + (24 \sqrt{2} + 16 \sqrt{2} - 24 \sqrt{2} - 16 \sqrt{2}) = 136 + 0 = 136 \]