Prove that the points (3, −2), (4, 0), (6, −3)

Question:

Prove that the points (3, −2), (4, 0), (6, −3) and (5, −5) are the vertices of a parallelogram.

Solution:

Let A (3,−2); B (4, 0); C (6,−3) and D (5,−5) be the vertices of a quadrilateral. We have to prove that the quadrilateral ABCD is a parallelogram.

We should proceed with the fact that if the diagonals of a quadrilateral bisect each other than the quadrilateral is a parallelogram.

Now to find the mid-point of two pointsand we use section formula as,

$\mathrm{P}(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

So the mid-point of the diagonal AC is,

$\mathrm{Q}(x, y)=\left(\frac{3+6}{2}, \frac{-2-3}{2}\right)$

$=\left(\frac{9}{2},-\frac{5}{2}\right)$

Similarly mid-point of diagonal BD is,

$R(x, y)=\left(\frac{4+5}{2}, \frac{-5+0}{2}\right)$

$=\left(\frac{9}{2},-\frac{5}{2}\right)$

Therefore the mid-points of the diagonals are coinciding and thus diagonal bisects each other.

Hence ABCD is a parallelogram.

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