Question

The line \((3a+1)x+(7a+2)y=17a+5\) represents concurrent lines. If \(d\) is distance from \((3,1)\) to line of slope 1 in this family, find \(2d^2\):

• A. 4
• B. 3
• C. 9
• D. 16

Hint:

First determine parameter using slope condition before distance formula.

Answer 1

The Correct Option is C

Concept: For slope 1 line, compare coefficients.

Step 1: Condition for slope 1.
\[ \frac{-(3a+1)}{7a+2}=1 \Rightarrow -(3a+1)=7a+2 \] \[ -3a-1=7a+2 \Rightarrow -10a=3 \Rightarrow a=-\frac{3}{10} \]

Step 2: Equation of line.
Substitute: \[ \left(-\frac{9}{10}+1\right)x+\left(-\frac{21}{10}+2\right)y= -\frac{51}{10}+5 \] Simplify: \[ \frac{1}{10}x - \frac{1}{10}y = -\frac{1}{10} \Rightarrow x-y+1=0 \]

Step 3: Distance from (3,1).
\[ d=\frac{|3-1+1|}{\sqrt{2}}=\frac{3}{\sqrt{2}} \] \[ 2d^2 = 2 \cdot \frac{9}{2}=9 \]