The equations of bisectors of two lines L1 L2 are 2 x-16 y-5=0 and 64 x+8 y+35=0. If the line L1 passes through (-11,4), the equation of acute angle bisector of L1 L2 is :

Question

The equations of bisectors of two lines L1 L2 are 2 x-16 y-5=0 and 64 x+8 y+35=0. If the line L1 passes through (-11,4), the equation of acute angle bisector of L1 L2 is : (A) 2 x-16 y-5=0 (B) 64 x+8 y+35=0 (C) data insufficient (D) none of these

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/249f129148cbf2b33e1bac1600da432c-.png" /> p=|(-22-64-5/√22+(-16)2)|=(91/√260) q=|(-64 × 11+8 × 4+35/√642+82)|=(637/2 √260) p < q Hence 2 x-16 y-5=0 is acute angle bisector

Q. The equations of bisectors of two lines $L_{1} & L_{2}$ are $2 x - 16 y - 5 = 0$ and $64 x + 8 y + 35 = 0$ . If the line $L_{1}$ passes through $(- 11, 4)$ , the equation of acute angle bisector of $L_{1} & L_{2}$ is :

$2 x - 16 y - 5 = 0$

$64 x + 8 y + 35 = 0$

data insufficient

none of these

$p = ∣ ∣ \frac{- 22 - 64 - 5}{2 ^{2} + ( - 16 ) ^{2}} ∣ ∣ = \frac{91}{260}$
$q = ∣ ∣ \frac{- 64 \times 11 + 8 \times 4 + 35}{6 4 ^{2} + 8 ^{2}} ∣ ∣ = \frac{637}{2 260}$
$p < q$ Hence $2 x - 16 y - 5 = 0$ is acute angle bisector

Q. The equations of bisectors of two lines L1​&L2​ are 2x−16y−5=0 and 64x+8y+35=0. If the line L1​ passes through (−11,4), the equation of acute angle bisector of L1​&L2​ is :

2x−16y−5=0

64x+8y+35=0