Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer.
(i) $(-1, -2), (1, 0), (-1, 2), (-3, 0)$
(ii) $(-3, 5), (3, 1), (0, 3), (-1, -4)$
(iii) $(4, 5), (7, 6), (4, 3), (1, 2)$

AcademicMathematicsNCERTClass 10

To do:

We have to find the quadrilateral formed, if any, by the given points.

Solution:

Let the given points be $A(-1, -2), B(1, 0), C(-1, 2), D(-3, 0)$.

We know that,

The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.

Therefore,

$\mathrm{AB}=\sqrt{(1+1)^{2}+(0+2)^{2}}$

Squaring on both sides, we get, $\mathrm{AB}^{2}=(1+1)^{2}+(0+2)^{2}$

$=(2)^{2}+(2)^{2}$

$=4+4$

$=8$

$\mathrm{BC}^{2}=(-1-1)^{2}+(2-0)^{2}$

$=(-2)^{2}+(2)^{2}$

$=4+4$

$=8$

$\mathrm{CD}^{2}=(-3+1)^{2}+(0-2)^{2}$

$=(-2)^{2}+(-2)^{2}$

$=4+4$

$=8$

$\mathrm{DA}^{2}=(-1+3)^{2}+(-2+0)^{2}$

$=(2)^{2}+(-2)^{2}$

$=4+4$

$=8$

$\mathrm{AC}^{2}=(-1+1)^{2}+(2+2)^{2}$

$=(0)^{2}+(4)^{2}$

$=0+16$

$=16$

$\mathrm{BD}^{2}=(-3-1)^{2}+(0-0)^{2}$

$=(-4)^{2}+0$

$=16$

Here,

$AB^2=BC^2=CD^2=DA^2$ and $AC^2=BD^2$

This implies,

$AB=BC=CD=DA$ and $AC=BD$.

All the sides are equal and the diagonals are equal to each other. 

Therefore, the quadrilateral formed by the points $A, B, C, D$ is a square.

(ii) Let the given points be $A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)$.

We know that,

The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.

Therefore,

$\mathrm{AB}=\sqrt{(3+3)^{2}+(1-5)^{2}}$

Squaring on both sides, we get,

$\mathrm{AB}^{2}=(3+3)^{2}+(1-5)^{2}$

$=(6)^{2}+(-4)^{2}$

$=36+16$

$=52$

$\mathrm{BC}^{2}=(0-3)^{2}+(3-1)^{2}$

$=(-3)^{2}+(2)^{2}$

$=9+4$

$=13$

$\mathrm{CD}^{2}=(-1-0)^{2}+(-4-3)^{2}$

$=(-1)^{2}+(-7)^{2}$

$=1+49$

$=50$

$\mathrm{DA}^{2}=(-3+1)^{2}+(5+4)^{2}$

$=(-2)^{2}+(9)^{2}$

$=4+81$

$=85$

$\mathrm{AC}^{2}=(0+3)^{2}+(3-5)^{2}$

$=(3)^{2}+ (-2)^{2}$

$=9+4$

$=13$

In $\Delta \mathrm{ABC}$,

$\mathrm{AB}=\sqrt{52}, \mathrm{AC}=\sqrt{13}, \mathrm{BC}=\sqrt{13}$

$\mathrm{AC}+\mathrm{BC}=\sqrt{13}+\sqrt{13}=2 \sqrt{13}$

$\mathrm{AB}=\sqrt{52}=2\sqrt{13}$

$\Rightarrow \mathrm{AC}+\mathrm{BC}=\mathrm{AB}$

This implies, the points $A, B ,C$ are collinear.

Therefore, $\Delta \mathrm{ABC}$ is not possible.

Hence, $\mathrm{ABCD}$ is not a quadrilateral.

(iii) Let the given points be $A (4, 5), B (7, 6), C (4, 3), D (1, 2)$.

We know that,

The distance between two points $\mathrm{A}(x_{1}, y_{1})$ and $\mathrm{B}(x_{2}, y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.

Therefore,

$\mathrm{AB}=\sqrt{(7-4)^{2}+(6-5)^{2}}$

Squaring on both sides, we get,

$\mathrm{AB}^{2}=(7-4)^{2}+(6-5)^{2}$

$=(3)^{2}+(1)^{2}$

$=9+1$

$=10$

$B C^{2}=(4-7)^{2}+(3-6)^{2}$

$=(3)^{2}+(-3)^{2}$

$=9+9$

$=18$

$\mathrm{CD}^{2}=(1-4)^{2}+(2-3)^{2}$

$=(-3)^{2}+(-1)^{2}$

$=9+1$

$=10$

$\mathrm{DA}^{2}=(4-1)^{2}+(5-2)^{2}$

$=(3)^{2}+(3)^{2}$

$=9+9$

$=18$

$\mathrm{AC}^{2}=(4-4)^{2}+(3-5)^{2}$

$=(0)^{2}+(-2)^{2}$

$=0+4$

$=4$

$\mathrm{BD}^{2}=(1-7)^{2}+(2-6)^{2}$

$=(-6)^{2}+(-4)^{2}$

$=36+16$

$=52$

Here,

$AB^2=CD^2$, $BC^2=DA^2$ and $AC^2≠BD^2$

This implies,

$A B=C D, B C=D A$ and $AC≠BD$

Opposite sides are equal and diagonals are not equal.

Therefore, the quadrilateral formed by the points $A, B, C, D$ is a parallelogram.

raja
Updated on 10-Oct-2022 13:22:03

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