Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram.

Question:

Show that A(1, 2), B(4, 3), C(6, 6) and D(3, 5) are the vertices of a parallelogram. Show that ABCD is not a rectangle.

Solution:

The given vertices are A(1, 2), B(4, 3), C(6, 6) and D(3, 5).

$A B=\sqrt{(1-4)^{2}+(2-3)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}}$

$=\sqrt{9+1}=\sqrt{10}$

$B C=\sqrt{(4-6)^{2}+(3-6)^{2}}=\sqrt{(-2)^{2}+(-3)^{2}}$

$=\sqrt{4+9}=\sqrt{13}$

$C D=\sqrt{(6-3)^{2}+(6-5)^{2}}=\sqrt{(3)^{2}+(1)^{2}}$

$=\sqrt{9+1}=\sqrt{10}$

$A D=\sqrt{(1-3)^{2}+(2-5)^{2}}=\sqrt{(-2)^{2}+(-3)^{2}}$

$=\sqrt{4+9}=\sqrt{13}$

$\because A B=C D=\sqrt{10}$ units and $B C=A D=\sqrt{13}$ units

Therefore, ABCD is a parallelogram. Now

$A C=\sqrt{(1-6)^{2}+(2-6)^{2}}=\sqrt{(-5)^{2}+(-4)^{2}}$

$=\sqrt{25+16}=\sqrt{41}$

$B D=\sqrt{(4-3)^{2}+(3-5)^{2}}=\sqrt{(1)^{2}+(-2)^{2}}$

$=\sqrt{1+4}=\sqrt{5}$

Thus, the diagonals AC and BD are not equal and hence ABCD is not a rectangle.

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