Question

Pair of lines through (1, 1) and making equal angle with 3x – 4y = 1 and 11x + 4y = 1 intersect x-axis at P1 and P2, then P1, P2 may be-

• A. $\left( \frac { 9 } { 7 } , 0 \right)$
• B. $\left( \frac { 7 } { 8 } , 0 \right)$ and (9, 0)
• C. $\left( \frac { 8 } { 7 } , 0 \right)$and $\left( \frac { 1 } { 8 } , 0 \right)$
• D. (8, 0) and $\left( \frac { 1 } { 8 } , 0 \right)$

Answer 1

Answer: (B)

$\left( \frac { 7 } { 8 } , 0 \right)$ and (9, 0)

Answer 2

Equation of bisectors $\frac { 3 x + 4 y - 1 } { 5 }$ = ± $\frac { 11 x + 4 y - 1 } { 15 }$

So bisectors have slope 8, – $\frac { 1 } { 8 }$

Equation of required lines y – 1 = 8(x – 1) and y – 1 = – $\frac { 1 } { 8 }$

(x – 1)

which intersect x-axis at $\left( \frac { 7 } { 8 } , 0 \right)$ and (9, 0).