# a quadrilateral has vertices at the points(-4,2) , (2,6), (8,5), (9,-7),show that the mid points of the sides of this quadrilateral are the vertices of a parallelogram.

The mid point of sides AB,BC,CD,DA are P,Q,R,S respectively.
And the mid-point formula is {(x1 + x2)/2 , (y1 + y2)/2}
So P is the midpoint of side AB
Hence P = {(-4+2)/2, (6+2)/2} = {-1, 4}

Q is the mid point of side BC
So Q = {(8+2)/2, (6+5)/2} = {5, 11/2}

R is the mid-point of CD
So R = {(8+9)/2, (5-7)/2} = {17/2, -1}

S is the mid-point of AD
So S = {(-4+9)/2, (2+ 7)/2} = {5/2, -5/2}

For the quadrilateral PQRS to be a parallelogram, the opposite sides are parallel, hence the slope of PQ = SR and slope of RQ = PS
So Now for finding the slope we have the formula m = $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$,
So slope of PQ = $\frac{\frac{11}{2}-4}{5-\left(-1\right)}=\frac{1}{4}$
Slope of RS = $\frac{\frac{-5}{2}-\left(-1\right)}{\frac{5}{2}-\frac{17}{2}}=\frac{1}{4}$
So slope of PQ = RS

And slope of RQ = $\frac{-1-\frac{11}{2}}{5-\frac{17}{2}}=\frac{\frac{-13}{2}}{\frac{-7}{2}}=\frac{13}{7}$
Slope of PS = $\frac{\frac{-5}{2}-4}{-1-\frac{5}{2}}=\frac{13}{7}$
Hence slope of RQ = PS
Hence PQRS is parallelogram

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