Lines: intersection, parallel, perpendicular and distance

Text

Given the lines $r:2x+3y-4=0$ and $s:y=-x+3$:

• (a) find the intersection of the two lines;
• (b) find the equation of the line through $B(-2;-2)$ parallel to $r$;
• (c) find the equation of the line through $B(-2;-2)$ perpendicular to $s$;
• (d) find the distance between $B$ and line $r$.

Solution

(a) Substituting $y=-x+3$ into $r$: $2x+3(-x+3)-4=0\Rightarrow -x+5=0\Rightarrow x=5$, so $y=-5+3=-2$. Intersection $\left(5;-2\right)$.

(b) Line $r$ has slope $m_r=-\tfrac23$. Parallel through $B(-2;-2)$: $y+2=-\tfrac23(x+2)\Rightarrow 2x+3y+10=0$.

(c) Line $s$ has slope $m_s=-1$; the perpendicular has slope $1$. Through $B(-2;-2)$: $y+2=1\cdot(x+2)\Rightarrow y=x$.

(d) Distance of $B(-2;-2)$ from line $r:2x+3y-4=0$: $d=\frac{|2(-2)+3(-2)-4|}{\sqrt{2^2+3^2}}=\frac{14}{\sqrt{13}}=\frac{14\sqrt{13}}{13}\approx 3{,}88.$

$\boxed{P\left(5;-2\right),\quad d=\frac{14\sqrt{13}}{13}\approx 3{,}88}$