The line L given by x5+ yb=1 passes through the point (13,32).

Question

The line $L$ given by $\dfrac x5+\dfrac yb=1$ passes through the point $(13,32)$. The line $K$ is parallel to $L$ and has the equation $\dfrac xc+\dfrac y3=1$. Then the distance between $L$ and $K$ is

Accepted Answer

Parallel slopes: $-\tfrac b5=-\tfrac3c\Rightarrow bc=15$. Since $(13,32)$ lies on $L$: $\tfrac{13}5+\tfrac{32}b=1\Rightarrow \tfrac{32}b=-\tfrac85\Rightarrow b=-20\Rightarrow c=-\tfrac34$.
Line $L$: $4x-y-20=0$. Line $K$ (from $x/c+y/3=1$ with $c=-3/4$): $4x-y+3=0$.
$$\text{Distance}=\frac{|-20-3|}{\sqrt{17}}=\frac{23}{\sqrt{17}}.$$

Solutions · 1