Line in fractional form: parallel, perpendicular and distance

Text

Given the line $r:\dfrac{2}{5}x+\dfrac{1}{5}y+1=0$:

• (a) find the intercepts with the axes;
• (b) decide whether $A(1;-3)$ lies on the line;
• (c) find the line $s$ perpendicular to $r$ through $A$;
• (d) find the line $t$ parallel to $r$ through the origin;
• (e) compute the distance between $A$ and line $r$.

Solution

Multiplying $r$ by $5$ gives $2x+y+5=0$, with slope $-2$.

(a) $x$-axis ($y=0$): $2x+5=0\Rightarrow x=-\tfrac52$, point $\left(-\tfrac52;0\right)$. $y$-axis ($x=0$): $y=-5$, point $(0;-5)$.

(b) $2(1)+(-3)+5=4\neq 0$: so $A$ does not lie on $r$.

(c) The perpendicular has slope $\tfrac12$. Through $A(1;-3)$: $y+3=\tfrac12(x-1)\Rightarrow x-2y-7=0$.

(d) Parallel to $r$ through the origin: $y=-2x$.

(e) Distance of $A(1;-3)$ from $2x+y+5=0$: $d=\frac{|2(1)+(-3)+5|}{\sqrt{2^2+1^2}}=\frac{4}{\sqrt5}=\frac{4\sqrt5}{5}\approx 1{,}79.$

$\boxed{\left(-\tfrac52;0\right),\ (0;-5);\quad d=\frac{4\sqrt5}{5}\approx 1{,}79}$