Question

If the points $A(0,4),B(1,2)$ and $C(3,3)$ are three corner of a square, find
the coordinates of the point at which the diagonals intersect.
the coordinates of $D$, the fourth corner of the square.

Answer 1

$A(0,4)B(1,2)C(3,3)$

Let $D$ be $(x,y)$

Slope of $AC=\frac{y_2-y_1}{x_2-x_1}=\frac{3-4}{3-0}=\frac{-1}{3}$

Slope of $BD=m_1{m}_2=-1$

$=-\frac{1}{3}{m}_2=-1$

$=m_2=3$

Midpoint of $BD=M.P$ of $AC$.

$(\frac{x+1}{2},\frac{y+2}{2})=(\frac{0+3}{2},\frac{4+3}{2})$

$\frac{x+1}{2}=\frac{3}{2}\to x+1=3$

$x=3-1$

$x=2$

$\frac{y+2}{2}=\frac{7}{2}$

$y+2=7$

$y=5$

$\therefore D=(2,5)$