$WXYZ$ is a square. If the co ordinates of $W$ and $Y$ are $(2,3)$ and $(5,7)$ respectively, find the co ordinates of $X$ and $Z$.
My Attempt:
I could only find the equation of diagonal $WY$ by using two point formula as follow $$(y-y_1)=\dfrac {y_2-y_1}{x_2-x_1} (x-x_1)$$ $$y-3=\dfrac {7-3}{5-2} (x-2)$$ $$4x-3y+1=0$$.
How do I complete the rest? please help
On
$$\begin{align} W(2,3)+(\;\;a,\;\;b)&=X(p,q)\qquad \cdots(1)\\ X(p,q)+(-b,\;\;a)&=Y(5,7)\qquad \cdots(2)\\ Y(5,7)+(-a,-b)&=Z(h,k)\qquad \cdots(3)\\ Z(h,k)+(\;\;b,-a)&=W(2,3)\;\;\;\quad \cdots(4)\end{align}$$ Adding $(1),(2)$ and solving gives $$(a,b)=\left(\frac 72, \frac 12\right)$$
Hence $$\color{red}{X=\left(\frac {11}2, \frac 72\right)\\ Z=\left(\frac 32, \frac {13}2\right)}$$ This satisfies $(3),(4)$ as well.
First find the length of diagonal and it's slope. And also it's midpoint Then you can find the slope of other diagonal as diagonals are perpendicular and they bisect each other.
Then using slope ,midpoint find eq of second diagonal and use the length half the diagonal to find the two other vertices