If ABC is an isosceles right angled ∆ at B and coordinates of A and C are (1,2) and (4,-2) then coordinates of Vertex B can be

Question

My approach : I have tried it by different methods but find them too lengthy like , by using Basic Distance formula . I am finding a short solution for it . A Hint will help.

Answer 1

The simplest way:

Find the midpoint $M$ of segment $AC$:

$$x_M=\frac{x_A+x_C}{2}=2.5\\y_M=\frac{y_A+y_C}{2}=0$$

Calculate:

$$\Delta x_C=x_C-x_M=1.5\\ \Delta y_C=y_C-y_M=-2$$

(Notice that $\Delta x_A=-\Delta x_C, \Delta y_A=-\Delta y_C$)

$MB$ is perpendicular to $MC$ and also $MB=MC$. It means that:

$$\Delta x_B=\mp \Delta y_C=\pm 2 \\ \Delta y_B=\pm\Delta x_C=\pm 1.5$$

So the coordinates of point $B$ are:

$$x_B=x_M + \Delta x_B=2.5 \pm 2 \\ y_B=y_M + \Delta y_B=0 \pm 1.5 $$

Finally, we have: $B(4.5,1.5)$ or $B(0.5,-1.5)$.

Answer 2

Solution with vectors.

We are looking for the vertices $B,B^{\,'}$ of the square $ABCB^{\,'}.$ Diagonals of the square are orthogonal, have equal length and bisect each other.

From $\vec{AC}(3,-4)$ we get $\vec{BB^{\,'}}(4,3)$ or $\vec{BB^{\,'}}(-4,-3).$ It suffices to continue with one of them, the second would give the same two points.

The midpoint $M\left(\frac{5}{2},0\right)$ of $AC$ is also midpoint of $BB^{\,'}.$

From this we get $B=M+\frac{1}{2}\vec{BB^{\,'}}$ and $B^{\,'}=M-\frac{1}{2}\vec{BB^{\,'}},$ or $B\left(\frac{9}{2},\frac{3}{2}\right)$ and $B^{\,'}\left(\frac{1}{2},-\frac{3}{2}\right).$