Question

The opposite angular points of a square are $(3,4)$ and $(1,-1)$. Find the co-ordinates of the other two vertices.

Answer 1

Length of diagonal $=\sqrt{2^2+5^2}=\sqrt{29}$

Side of square $=\frac{\sqrt{29}}{\sqrt{2}}=\sqrt{\frac{29}{2}}$

Angle between diagonal and side is ${45}^{\circ }$

Slope of diagonal $=\frac{4+1}{3-1}=\frac{5}{2}$

$\tan {45}^{\circ }=\frac{m_1-\frac{5}{2}}{1+\frac{5}{2}{m}_1}$ $1+\frac{5}{2}{m}_1=m_1-\frac{5}{2}$ $m_1=-\frac{7}{3}$

Slope of arc side $=-\frac{7}{3}$ slope of others side $=\frac{3}{7}$

$\frac{4-y_2}{3-x_2}=\frac{-7}{3}$ ____ (1)

$\frac{4-y_1}{3-x_1}=\frac{3}{7}$ ______(2)

$\frac{29}{2}=(3-x_1{)}^2+(4-y_1{)}^2$ _______ (3)

$\frac{29}{2}=(3-x_2{)}^2+(4-x_2{)}^2$ _____ (4)

Solve (1),(2),(3),(4)

Points are $(x_1,y_1)=(\frac{-1}{2},\frac{5}{2})$

$(x_2,y_2)=(\frac{3}{2},\frac{15}{2})$