Name the quadrilateral formed, if any, by the following points, and given reasons for your answers:
(i) A(−1,−2) B(1, 0), C (−1, 2), D(−3, 0)
(ii) A(−3, 5) B(3, 1), C (0, 3), D(−1, −4)
(iii) A(4, 5) B(7, 6), C (4, 3), D(1, 2)
(i) A (−1,−2) , B(1,0), C(−1,2), D(−3,0)
Let A, B, C and D be the four vertices of the quadrilateral ABCD.
We know the distance between two points Pand Qis given by distance formula:
$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Hence
$\Rightarrow A B=\sqrt{(1-(-1))^{2}+(0-(-2))^{2}}$
$\Rightarrow A B=\sqrt{(2)^{2}+(2)^{2}}$
$\Rightarrow A B=\sqrt{4+4}$
$\Rightarrow A B=\sqrt{4}$
$\Rightarrow A B=2 \sqrt{2}$
Similarly,
$\Rightarrow B C=\sqrt{((-1)-1)^{2}+(2-0)^{2}}$
$\Rightarrow B C=\sqrt{(-2)^{2}+(2)^{2}}$
$\Rightarrow B C=\sqrt{4+4}$
$\Rightarrow B C=\sqrt{8}$
$\Rightarrow B C=2 \sqrt{2}$
Similarly,
$\Rightarrow C D=\sqrt{((-3)-(-1))^{2}+(0-(2))^{2}}$
$\Rightarrow C D=\sqrt{(-2)^{2}+(-2)^{2}}$
$\Rightarrow C D=\sqrt{4+4}$
$\Rightarrow C D=\sqrt{8}$
$\Rightarrow C D=2 \sqrt{2}$
Also,
$\Rightarrow D A=\sqrt{((-1)-(-3))^{2}+(0-(-2))^{2}}$
$\Rightarrow D A=\sqrt{(2)^{2}+(2)^{2}}$
$\Rightarrow D A=\sqrt{4+4}$
$\Rightarrow D A=\sqrt{8}$
$\Rightarrow D A=2 \sqrt{2}$
Hence from above we see that all the sides of the quadrilateral are equal. Hence it is a square.
(ii) A (−3,5) , B(3,1), C(0,3), D(−1,−4)
Let A, B, C and D be the four vertices of the quadrilateral ABCD.
We know the distance between two points Pand Qis given by distance formula:
$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Hence
$\Rightarrow A B=\sqrt{(3-(-3))^{2}+(1-(5))^{2}}$
$\Rightarrow A B=\sqrt{(6)^{2}+(4)^{2}}$
$\Rightarrow A B=\sqrt{36+16}$
$\Rightarrow A B=\sqrt{52}$
$\Rightarrow A B=2 \sqrt{13}$
Similarly,
$\Rightarrow B C=\sqrt{(0-3)^{2}+(3-1)^{2}}$
$\Rightarrow B C=\sqrt{(-3)^{2}+(2)^{2}}$
$\Rightarrow B C=\sqrt{9+4}$
$\Rightarrow B C=\sqrt{13}$
Similarly,
$\Rightarrow C D=\sqrt{((-1)-0)^{2}+((-4)-(3))^{2}}$
$\Rightarrow C D=\sqrt{(-1)^{2}+(-7)^{2}}$
$\Rightarrow C D=\sqrt{1+49}$
$\Rightarrow C D=\sqrt{50}$
$\Rightarrow C D=5 \sqrt{2}$
Also,
$\Rightarrow D A=\sqrt{((-1)-(-3))^{2}+((-4)-5)^{2}}$
$\Rightarrow D A=\sqrt{(2)^{2}+(-9)^{2}}$
$\Rightarrow D A=\sqrt{4+81}$
$\Rightarrow D A=\sqrt{85}$
(iii) A (4, 5), B (7,6), C(4,3), D(1,2)
Let A, B, C and D be the four vertices of the quadrilateral ABCD.
We know the distance between two points Pand Qis given by distance formula:
$\mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$
Hence
$\Rightarrow A B=\sqrt{(7-4)^{2}+(6-5)^{2}}$
$\Rightarrow A B=\sqrt{(3)^{2}+(1)^{2}}$
$\Rightarrow A B=\sqrt{9+1}$
$\Rightarrow A B=\sqrt{10}$
Similarly,
$\Rightarrow B C=\sqrt{(4-7)^{2}+(3-6)^{2}}$
$\Rightarrow B C=\sqrt{(-3)^{2}+(-3)^{2}}$
$\Rightarrow B C=\sqrt{9+9}$
$\Rightarrow B C=\sqrt{18}$
Similarly,
$\Rightarrow C D=\sqrt{(1-4)^{2}+(2-3)^{2}}$
$\Rightarrow C D=\sqrt{(-3)^{2}+(-1)^{2}}$
$\Rightarrow C D=\sqrt{9+1}$
$\Rightarrow C D=\sqrt{10}$
Also,
$\Rightarrow D A=\sqrt{(1-4)^{2}+(2-5)^{2}}$
$\Rightarrow D A=\sqrt{(-3)^{2}+(-3)^{2}}$
$\Rightarrow D A=\sqrt{9+9}$
$\Rightarrow D A=\sqrt{18}$
Hence from above we see that
AB = CD and BC = DA
Here opposite sides of the quadrilateral is equal. Hence it is a parallelogram.
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