Question

Prove that the points (2,-2),(-2,1),(5,2) are the vertices of the right angle triangle also find area of the triangle

Answer 1

1st use distance formula for each side of the triangle
then use PT for checking if they are equal then u will get it as equal
then use area of triangle formula u will get the ans

Answer 2

Answer:

$\orange {ABC \:is \:a \: right \: triangle .}$

$\red {Area \: of \: \triangle ABC }$

$\green {= 12.5 \: square \: units}$

Step-by-step explanation:

$Let \:A(2,2)=(x_{1},y_{1}),\:\:B(-2,1)=(x_{2},y_{2}),\:</p><p>\\and \:C(5,2)=(x_{3},y_{3}),\:are \: vertices \:of \\a \: triangle$

$\blue { Distance \: between \:A\:and \: B}$

$\orange { = \sqrt{(x_{2}-x_{1})^{2} + (y_{2} - y_{1})^{2}}}$

$= \sqrt{(-2-2)^{2} + [1 - (-2]^{2}}$

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

$= \sqrt{ 16 + 9 }$

$= \sqrt{25}$

$= 5 \: ---(1)$

$\blue { Distance \: between \:B\:and \: C}$

$= \sqrt{[5-(-2)]^{2} + (2- 1)^{2}}$

$= \sqrt{7^{2} + 1^{2}}$

$= \sqrt{ 49 + 1 }$

$= \sqrt{50}$

$= \sqrt{2\times 5^{2}}\\=5\sqrt{2}}---(2)$

$\blue { Distance \: between \:C\:and \: A}$

$= \sqrt([5-2)^{2} + [2-(-2)]^{2}}$

$= \sqrt{3^{2} + 4^{2}}$

$= \sqrt{ 9 + 16 }$

$= \sqrt{25}$

$= 5---(3)$

$AB^{2} = 25$

$BC^{2} = 50$

$CA^{2} = 25$

$BC^{2} = AB^{2} + CA^{2}$

$\orange {ABC \:is \:a \: right \: triangle .}$

$\blue {\underline \:By \: Pythagoras\: Theorem }$

$\red {Area \: of \: \triangle ABC }$

$=\frac{1}{2}\times AB \times CA$

$= \frac{1}{2}\times 5 \times 5$

$= \frac{25}{2} \\= 12.5 \: square \: units$

Therefore.,

$\orange {ABC \:is \:a \: right \: triangle .}$

$\red {Area \: of \: \triangle ABC }$

$\green {= 12.5 \: square \: units}$

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