All possible values of coordinate k such that triangle ABC is a right triangle?

Question

Determine all values of $k$ for which the points $A = (1,2),$ $B = (11,2),$ and $C = (k,6)$ form the vertices of a right-angled triangle.

Answer 1

Hint:

If the right angle is at $A$, $k=1$.

If the right angle is at $B$, $k=11$.

If the right angle is at $C$, then $AC\perp BC$. We can find $k$ easily by considering the product of the slopes.

Answer 2

Hint:

a) If $\triangle ABC$ is riht-angled at $A$, then $AC$ is vertical, i.e. $k=1$.

b) If $\triangle ABC$ is riht-angled at $B$, then $BC$ is vertical, i.e. $k=11$.

c) If $\triangle ABC$ is riht-angled at $C$, then $m_{BC}m_{CA}=-1$.

Answer 3

we have $$AB^2=100,BC^2=(k-11)^2+4^2,AC^2=(k-1)^2+4^2$$ and solving the equation $$AB^2=AC^2+BC^2$$ for $k$ we get $$k_1=3$$ or $$k_2=9$$ can you proceed?

Answer 4

$\vec{AB}=(10,0)$ and we know that either $\vec{AC}=(k-1,4)$ or $\vec{BC}=(k-11,4)$ must be orthogonal to $\vec{AB}$, or that $\vec{AC}$ and $\vec{BC}$ are orthogonal.

$0=\langle(10,0),(k-1,4)\rangle=10(k-1)$ iff $k=1$ and

$0=\langle(10,0),(k-11,4)\rangle=10(k-11)$ iff $k=11$ and

$0=\langle(k-1,4),(k-11,4)\rangle=k^2-12k+11+16=(k-3)(k-9)=0$ iff $k=3$ or $k=9$.

So $k$ is either $1$, $3$, $9$ or $11$.

Answer 5

If the right angle is at $A$, then $k=1$. If the right angle is at $B$, then $k=11$. If the right angle is at $C$, then the product of the slope of $AC$ and $BC$, is $-1$, or, $$\frac{6-2}{k-1}\cdot\frac{6-2}{k-11}=-1$$ Sobe this quadratic in $k$ to get the values of $k$ for which the right angle is at $C$.