Right triangle in the Cartesian plane

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Given the triangle with vertices $A(0;1)$, $B(1;4)$, $C(7;2)$:

• (a) find the side lengths and verify the triangle is right-angled;
• (b) find the equation of line $AC$;
• (c) find the length of the altitude to the hypotenuse using the point–line distance;
• (d) find the line through $B$ perpendicular to $AC$ and its intersection point $H$ with $AC$.

Solution

(a) $\overline{AB}=\sqrt{1^2+3^2}=\sqrt{10}$, $\overline{BC}=\sqrt{6^2+(-2)^2}=\sqrt{40}=2\sqrt{10}$, $\overline{AC}=\sqrt{7^2+1^2}=\sqrt{50}=5\sqrt2$. Since $\overline{AB}^2+\overline{BC}^2=10+40=50=\overline{AC}^2$, the triangle is right-angled at $B$.

(b) Line $AC$: slope $\dfrac{2-1}{7-0}=\dfrac17$, so $y=\dfrac{x}{7}+1$, i.e. $x-7y+7=0$.

(c) Altitude $=$ distance of $B(1;4)$ from $AC$: $h=\frac{|1-7\cdot 4+7|}{\sqrt{1^2+7^2}}=\frac{20}{5\sqrt2}=2\sqrt2\approx 2{,}828.$

(d) The perpendicular through $B$ has slope $-7$: $y-4=-7(x-1)\implies y=-7x+11$. Intersection with $AC$: $\frac{x}{7}+1=-7x+11 \implies 50x=70 \implies x=\tfrac75,\ y=\tfrac65.$ Hence $H\left(\tfrac75;\tfrac65\right)=(1{,}4;\,1{,}2)$.

$\boxed{\overline{AB}=\sqrt{10},\ \overline{BC}=2\sqrt{10},\ \overline{AC}=5\sqrt2,\ h=2\sqrt2,\ H\left(\tfrac75;\tfrac65\right)}$