Question

show that the point (3,-2),(7,6),(-1,2),(-5,-6) form a rhombus
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Answer 1

Answer:

Let rhombus MNPQ has point M(3,-2),N(7,6),P(-1,2),Q(-5,-6).

We know that rhombus is a quadrilateral has all equal sides and diagonal cuts each other equally at 90°.

In Cartesian plane by distance formula, MN=√(6+2)^2+(7–3)^2 =√64+16=√80.

NP=√(-1–7)^2+(2–6)^2=√80.

PQ=√(-5–(-1))^2+(-6–2)^2 =√16+64=√80.

MQ=√(-5–3)^2+(-6-(-2))^2=√64+16=√80.

So all sides are equal.

Diagonal MP=√(-1–3)^2+(2-(-2)^2)=√16+16=√32=4√2

NQ=√(-5–7)^2+(-6–6)^2=√144+144=√288=12√2

Now MP divides NQ equally so midean of diagonals NQ=6√2 and MP=4√2.

These diagonals divide each other orthogonal or at 90° so,

Side=√(6√2)^2+(2√2)^2 =√72+8=√80.

This shows diagonals divide each other at 90° by converse of Pythagorean theorem that √(base)^2+(perpendicular)^2= hypotenuse then two sides cuts at 90°.

Above verification shows that given points form a rhombus.

Step-by-step explanation:

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