Question

A (1,-2) B (5,2) C(3,-1)
D(-1,-5) are any four points.
Frame one mathematically correct
question using all four points and
solve it.​

Answer 1

$\red {Show \:that \: the \: points \:A(1,-2) , }$

$\red{B(5,2), C(3,-1) \:and \: D(-1,-5) \:taken \:in}$

$\red{ that \:order \:are \: Vertices \:of \:a }$ $\red {parallelogram.}$

$\underline { \green { Solution :}}$

Let the points A(1,-2), B(5,2) , C(3,-1) and D(-1,-5)

are Vertices of a parallelogram.

We know that the diagonals of a parallelogram bisects each other .

Therefore,

The midpoint of the diagonals AC and DB should be same .

$Now, we\: find \:the \: mid-points \:of \:AC \\and \:DB \:by \:using \\ \big( \frac{x_{1}+x_{2}}{2},\\frac{y_{1}+y_{2}}{2}\big) \: formula .$

$Mid-point \:of \:AC = \big( \frac{1+3}{2},\frac{-2-1}{2}\big) \\= \big( \frac{4}{2}, \frac{-3}{2}\big) \\= \big( 2 , \frac{-3}{2}\big) \: --(1)$

$Mid-point \:of \:DB = \big( \frac{5-1}{2},\frac{2-5}{2}\big) \\= \big( \frac{4}{2}, \frac{-3}{2}\big) \\= \big( 2 , \frac{-3}{2}\big) \: --(2)$

$Hence, midpoint \:of \:AC \:and \:midpoint \\of \:DB \:is \:same .$

Therefore.,

$The \:points \:A,B,C\:and \:D \:are \: Vertices \\of \:a \: parallelogram.$

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