Question

The vertices of a quadrilateral, taken in order, are the points $(0,0)$, $(4,0)$, $(6,7)$, and $(0,3)$; find the coordinates of the point of intersection of the two lines joining the middle points of opposite sides.

Answer 1

$A(0,0),B(4,0),C(6,7),D(0,3)$

Mid point of $AB$

$P(\frac{0+4}{2},\frac{0+0}{2})\Rightarrow P(2,0)$

Mid point of $BC$

$Q(\frac{4+6}{2},\frac{0+7}{2})\Rightarrow Q(5,\frac{7}{2})$

Mid point of $CD$

$R(\frac{6+0}{2},\frac{7+3}{2})\Rightarrow R(3,5)$

Mid point of $DA$

$S(\frac{0+0}{2},\frac{3+0}{2})\Rightarrow S(0,\frac{3}{2})$

Equation of line joining $PR$

$y-0=\frac{5-0}{3-2}(x-2)y-0=5x-105x-y=\mathrm{10........}\left(i\right)$

Equation of $QS$

$y-\frac{7}{2}=\frac{\frac{3}{2}-\frac{7}{2}}{0-5}(x-5)y-\frac{7}{2}=\frac{2}{5}(x-5)5(2y-7)=4(x-5)10y-35=4x-2010y-4x=\mathrm{15......}\left(ii\right)$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow x=\frac{5}{2},y=\frac{5}{2}$

So the point of intersection is $(\frac{5}{2},\frac{5}{2})$