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

Question

Let the point $P(\alpha,\beta)$ be at a unit distance from each of the two lines $L_1: 3x-4y+12=0$, and $L_2: 8x+6y+11=0$. If $P$ lies below $L_1$ and above $L_2$, then $100(\alpha+\beta)$ is equal to:

Accepted Answer

$P$ lies on the bisector of the angle between $L_1,L_2$ containing the region below $L_1$ and above $L_2$:
$$2(3x-4y+12) = 8x+6y+11 \implies 2x+14y-13=0 \quad \text{(i)}$$

Also $P$ at unit distance from $L_1$: $\dfrac{3\alpha-4\beta+12}5=1 \implies 3\alpha-4\beta+7=0 \quad \text{(ii)}$

Solving (i) and (ii): $\alpha=-\dfrac{23}{25},\ \beta=\dfrac{53}{50}$.

$$100(\alpha+\beta) = 100\left(-\frac{23}{25}+\frac{53}{50}\right) = 100\left(\frac{-46+53}{50}\right)=100\cdot\frac7{50}=14$$

Solutions · 1