# The points A(4, 3), B(6, 4), C(5, – 6)

Question:

The points A(4, 3), B(6, 4), C(5, – 6) and D(- 3, 5) are vertices of a parallelogram.

Solution:

False

Now, distance between $A(4,3)$ and $B(6,4), A B=\sqrt{(6-4)^{2}+(4-3)^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$

$\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

Distance between $B(6,4)$ and $C(5,-6), B C=\sqrt{(5-6)^{2}+(-6-4)^{2}}$

$=\sqrt{(-1)^{2}+(-10)^{2}}$

$=\sqrt{1+100}=\sqrt{101}$

Distance between $C(5,-6)$ and $D(-3,5), C D=\sqrt{(-3-5)^{2}+(5+6)^{2}}$

$=\sqrt{(-8)^{2}+11^{2}}$

$=\sqrt{64+121}=\sqrt{185}$

Distance between $D(-3,5)$ and $A(4,3), D A=\sqrt{(4+3)^{2}+(3-5)^{2}}$

$=\sqrt{7^{2}+(-2)^{2}}$

$=\sqrt{49+4}=\sqrt{53}$

In parallelogram, opposite sides are equal. Here, we see that all sides AB, BC, CD and DA are different.

Hence, given vertices are not the vertices of a parallelogram.