Question

plot the following plots on the graph and identify the type of figure, A) (-2,4); B) (2,0) C) (-6,0) D) (-2,-4)

Answer 1

$\sf\huge\bold{Answer:}$

$\fbox{\rm Square}$

Given :

Figure ACDB

Points :

A (-2,4)

B(2,0)

C(-6,0)

D (-2,-4)

To find :

Figure formed by plotting points on graph

Solution :

Figure is in attachment.

• Distance formula :

$\tt \boxed{\bf d = \sqrt{(x_2 - x_1) {}^{2} + (y_2 -y_1 ) {}^{2} }}$

where,

$\tt x_2$ = x coordinate of 2nd point.

$\tt x_1$ = x coordinate of 1st point.

$\tt y_2$ = y coordinate of 2nd point.

$\tt y_1$ = y coordinate of 1st point.

• To find AC

→A(-2,4)

→C(-6,0)

$\tt AC = \sqrt{(-6 - (-2)) {}^{2} + (0 -4 ) {}^{2} }$

$\tt AC = \sqrt{(-6+2) {}^{2} + ( -4 ) {}^{2} }$

$\tt AC = \sqrt{(-4) {}^{2} + (-4 ) {}^{2} }$

$\tt AC = \sqrt{(16+ 16 }$

$\tt AC = \sqrt{32 }$

$\tt\red{ AC = 4\sqrt{ 2 }\:units}$

• To find CD

→C(-6,0)

→D(-2,-4)

$\tt CD = \sqrt{(-2-( - 6) ) {}^{2} + (-4-0 ) {}^{2} }$

$\tt CD = \sqrt{(-2+6) {}^{2} + (-4 ) {}^{2} }$

$\tt CD = \sqrt{(4) {}^{2} + (-4 ) {}^{2} }$

$\tt CD = \sqrt{16+ 16 }$

$\tt CD = \sqrt{32 }$

$\tt \red{CD = 4\sqrt{2 }\:units}$

• To find BD

→B(2,0)

→D(-2,-4)

$\tt BD = \sqrt{(-2 - 2 ) {}^{2} + (-4 -0 ) {}^{2} }$

$\tt BD = \sqrt{(-4 ) {}^{2} + (-4 ) {}^{2} }$

$\tt BD = \sqrt{16 + 16 }$

$\tt BD = \sqrt{32 }$

$\tt\red{ BD = 4\sqrt{2} {\:units }}$

• To find AB

→A(-2,4)

→B(2,0)

$\tt AB = \sqrt{(2 - (-2)) {}^{2} + (0 -4 ) {}^{2} }$

$\tt AB = \sqrt{(2 +2) {}^{2} + ( -4 ) {}^{2} }$

$\tt AB = \sqrt{(4) {}^{2} + (-4 ) {}^{2} }$

$\tt AB = \sqrt{(16+ 16 }$

$\tt AB = \sqrt{32 }$

$\tt\red{ AB = 4\sqrt{ 2 }\:units}$

Hence, it can be seen that all sides of figure are equal.

Now, let us see whether diagonals are equal or not.

• To find BC

→B(2,0)

→C(-6,0)

$\tt BC = \sqrt{(-6 - 2 ) {}^{2} + (0 -0 ) {}^{2} }$

$\tt BC = \sqrt{(-8 ) {}^{2} + (0 ) {}^{2} }$

$\tt BC = \sqrt{64 + 0 }$

$\tt BC = \sqrt{64 }$

$\tt\red{ BC = {8\:units }}$

• To find DA

→D(-2,-4)

→A(-2,4)

$\tt DA = \sqrt{(-2-( - 2) ) {}^{2} + (4-(-4) ) {}^{2} }$

$\tt DA = \sqrt{(-2+ 2 ) {}^{2} + (4+4 ) {}^{2} }$

$\tt DA = \sqrt{(0) {}^{2} + (8 ) {}^{2} }$

$\tt DA = \sqrt{0 + 64 }$

$\tt\red{ DA = 8\:units}$

Diagonals are also equal.

Hence, a figure having all four sides and 2 of it's diagonal equal is a square.

The figure formed is a square.