Question

distance formula

Let the four vertices be A(1,1),B(-1,5),C(7,9),D(9,5). Find the sides AB, BC, CD, DA and the 2 diagonals AC and BD. Based
on the measurements tell what type of Quadrilateral is it.
Do it on a paper and upload here.
Your work
Assigned​

Answer 1

Answer:–

Given,

• A (1, 1)
• B (-1, 5)
• C (7, 9)
• D (9, 5)

To Find,

• Distance AB
• Distance BC
• Distance CD
• Distance AD
• Distance AC (Diagonal)
• Distance BD (Diagonal)
• What type of Quadrilateral it is?

Formula,

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

Solution –

• Distance AB

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

$\leadsto\sf{\sqrt{{(-1-1)}^{2}+{(5-1)}^{2}}}$

$\leadsto\sf{\sqrt{{(-2)}^{2}+{(4)}^{2}}}$

$\leadsto\sf{\sqrt{4+16}}$

$\leadsto\sf\underline{\bf{\sqrt{20}=2\sqrt{5}\;units}}$

$\rule{100}1$

• Distance BC

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

$\leadsto\sf{\sqrt{{(7-(-1))}^{2}+{(9-5)}^{2}}}$

$\leadsto\sf{\sqrt{{(8)}^{2}+{(4)}^{2}}}$

$\leadsto\sf{\sqrt{64+16}}$

$\leadsto\sf\underline{\bf{\sqrt{80}=4\sqrt{5}\:units}}$

$\rule{100}1$

• Distance CD

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

$\leadsto\sf{\sqrt{{(9-7)}^{2}+{(5-9)}^{2}}}$

$\leadsto\sf{\sqrt{{(2)}^{2}+{(-4)}^{2}}}$

$\leadsto\sf{\sqrt{4+16}}$

$\leadsto\sf\underline{\bf{\sqrt{20}=2\sqrt{5}\:units}}$

$\rule{100}1$

• Distance DA

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

$\leadsto\sf{\sqrt{{(9-1)}^{2}+{(5-1)}^{2}}}$

$\leadsto\sf{\sqrt{{(8)}^{2}+{(4)}^{2}}}$

$\leadsto\sf{\sqrt{64+16}}$

$\leadsto\sf\underline{\bf{\sqrt{80}=4\sqrt{5}\:units}}$

$\rule{100}1$

• Diagonal AC

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

$\leadsto\sf{\sqrt{{(7-1))}^{2}+{(9-1)}^{2}}}$

$\leadsto\sf{\sqrt{{(6)}^{2}+{(8)}^{2}}}$

$\leadsto\sf{\sqrt{36+64}}$

$\leadsto\sf\underline{\bf{\sqrt{100}=10\:units}}$

$\rule{100}1$

• Diagonal BD

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

$\leadsto\sf{\sqrt{{(9-(-1))}^{2}+{(5-5)}^{2}}}$

$\leadsto\sf{\sqrt{{(10)}^{2}+{(0)}^{2}}}$

$\leadsto\sf{\sqrt{100+0}}$

$\leadsto\sf\underline{\bf{\sqrt{100}=100\:units}}$

$\rule{100}1$

Hence, we observe :-

• Opposite sides equal
• Diagonals equal

So, the quadrilateral is a Rectangle.

$\rule{100}2$