Let A be the point of intersection of the lines 3x+2y=14,5x-y=6 and B be the point of inte…

Question

Let $A$ be the point of intersection of the lines $3x+2y=14,5x-y=6$ and $B$ be the point of intersection of the lines $4x+3y=8,6x+y=5$. The distance of the point $P(5,-2)$ from the line $AB$ is

Accepted Answer

Solving $3x+2y=14$ and $5x-y=6$: from the second, $y=5x-6$; substitute: $3x+2(5x-6)=14 \implies 13x=26\implies x=2,\ y=4$. So $A=(2,4)$.

Solving $4x+3y=8$ and $6x+y=5$: from the second, $y=5-6x$; substitute: $4x+3(5-6x)=8 \implies -14x=-7\implies x=\dfrac12,\ y=2$. So $B=\left(\dfrac12,2\right)$.

Line $AB$: slope $=\dfrac{4-2}{2-\frac12}=\dfrac{2}{3/2}=\dfrac43$, so $AB: y-4=\dfrac43(x-2) \implies 4x-3y+4=0$.

$$\text{Distance from }P(5,-2) = \frac{|20+6+4|}{\sqrt{16+9}} = \frac{30}{5}=6$$

Solutions · 1