Two lines: forms, parallel, perpendicular and distance

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Given the lines $r:-\dfrac{4}{3}x-\dfrac{2}{5}y+1=0$ and $s:y=-\dfrac{1}{2}x+\dfrac{3}{2}$:

• (a) find the intersection of the two lines;
• (b) write both in implicit and explicit form;
• (c) find the line through $B(-8;1)$ parallel to $r$;
• (d) find the line through $C(-4;-1)$ perpendicular to $s$;
• (e) find the distance between $B$ and line $r$.

Solution

(b) It is convenient to start from the forms. From $r:-\tfrac43x-\tfrac25y+1=0$, multiplying by $-15$ gives the implicit form $20x+6y-15=0$; solving for $y$, $y=\tfrac52-\tfrac{10}{3}x$, with slope $m_r=-\tfrac{10}{3}$. The line $s:y=-\tfrac12x+\tfrac32$ is already explicit and has implicit form $x+2y-3=0$, slope $m_s=-\tfrac12$.

(a) Equating the explicit forms: $-\tfrac{10}{3}x+\tfrac52=-\tfrac12x+\tfrac32$. Multiplying by $6$: $-20x+15=-3x+9\Rightarrow -17x=-6\Rightarrow x=\tfrac{6}{17}$. Then $y=\tfrac32-\tfrac12\cdot\tfrac{6}{17}=\tfrac{51-6}{34}=\tfrac{45}{34}$. Intersection $\left(\tfrac{6}{17};\tfrac{45}{34}\right)\approx(0{,}35;1{,}32)$.

(c) Parallel to $r$ (slope $-\tfrac{10}{3}$) through $B(-8;1)$: $y-1=-\tfrac{10}{3}(x+8)\Rightarrow 10x+3y+77=0$.

(d) The perpendicular to $s$ has slope $2$. Through $C(-4;-1)$: $y+1=2(x+4)\Rightarrow y=2x+7$.

(e) Distance of $B(-8;1)$ from $r:20x+6y-15=0$: $d=\frac{|20(-8)+6(1)-15|}{\sqrt{20^2+6^2}}=\frac{169}{\sqrt{436}}=\frac{169}{2\sqrt{109}}\approx 8{,}09.$

$\boxed{P\left(\tfrac{6}{17};\tfrac{45}{34}\right)\approx(0{,}35;1{,}32),\quad d=\frac{169}{2\sqrt{109}}\approx 8{,}09}$