Let the point [P(, )] be at a unit distance from each of the two lines L_1: 3 x-4 y+12=0, …

Question

Let the point $$P(\alpha, \beta)$$ be at a unit distance from each of the two lines $L_{1}: 3 x-4 y+12=0$, and $L_{2}: 8 x+6 y+11=0$. If $P$ lies below $L_{1}$ and above ${ }{L_{2}}$, then $$100(\alpha+\beta)$$ is equal to :

Accepted Answer

${L_1}:3x - 4y + 12 = 0$

${L_2}:8x + 6y + 11 = 0$

Equation of angle bisector of L1 and L2 of angle containing origin

$2(3x - 4y + 12) = 8x + 6y + 11$

$2x + 14y - 13 = 0$ ...... (i)

$${{3\alpha  - 4\beta  + 12} \over 5} = 1$$

$$ \Rightarrow 3\alpha  - 4\beta  + 7 = 0$$ ...... (ii)

Solution of $2x + 14y - 13 = 0$ and $3x - 4y + 7 = 0$ gives the required point $$P(\alpha ,\beta ),\,\alpha  = {{ - 23} \over {25}},\beta  = {{53} \over {50}}$$

$$100(\alpha  + \beta ) = 14$$

Solutions · 1