Bisector, angle bisectors and distance in a triangle

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Consider the points $A(3;1)$, $B(-1;4)$, $C(-2;-1)$:

• (a) find the perpendicular bisector of segment $AC$;
• (b) find the lines $AB$ and $BC$;
• (c) find the bisectors of the angles formed by those two lines;
• (d) find the distance between $B$ and line $AC$.

Solution

(a) Midpoint of $AC$: $\left(\tfrac12;0\right)$; slope $AC=\tfrac25$, perpendicular $-\tfrac52$: bisector $y=-\tfrac52 x+\tfrac54$.

(b) $AB$: slope $-\tfrac34$, $3x+4y-13=0$. $BC$: slope $5$, $5x-y+9=0$.

(c) Angle bisectors: equate the distances from the two lines: $\frac{3x+4y-13}{5}=\pm\frac{5x-y+9}{\sqrt{26}}.$

(d) Line $AC$: $2x-5y-1=0$; distance of $B(-1;4)$: $d=\frac{|2(-1)-5(4)-1|}{\sqrt{29}}=\frac{23}{\sqrt{29}}\approx 4{,}27.$

$\boxed{d(B,\,r_{AC})=\frac{23}{\sqrt{29}}\approx 4{,}27}$