Line: axis intercepts, parallel and perpendicular

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Given the line $r:2x+y+1=0$:

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

Solution

(a) With the $x$-axis ($y=0$): $2x+1=0\Rightarrow x=-\tfrac12$, point $\left(-\tfrac12;0\right)$. With the $y$-axis ($x=0$): $y=-1$, point $(0;-1)$.

(b) $2(3)+3+1=10\neq 0$: so $A$ does not lie on $r$.

(c) Line $r$ has slope $-2$; the perpendicular has slope $\tfrac12$. Through $A(3;3)$: $y-3=\tfrac12(x-3)\Rightarrow x-2y+3=0$.

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

(e) Distance of $A(3;3)$ from $r$: $d=\frac{|2(3)+3+1|}{\sqrt{2^2+1^2}}=\frac{10}{\sqrt5}=2\sqrt5\approx 4{,}47.$

$\boxed{\left(-\tfrac12;0\right),\ (0;-1);\quad d=2\sqrt5\approx 4{,}47}$