Question

$\huge\red{\boxed{\blue{\mathcal{\overbrace{\underbrace{\fcolorbox{blue}{aqua}{\underline{\red{QUESTIONS}}}}}}}}}$


Prove that the points (7, 10), (-2, 5) and (3, -4) are the vertices of an isosceles right triangle



➳ REQUIRED GOOD ANSWER​

Answer 1

Let A(7, 10), B(-2, 5), C(3, -4) be vertices of an isoceles triangle.

$\small\tt\pink{Then \: AB}= \sqrt{(7 + 2)} {}^{2} + \sqrt{(10 - 5) {}^{2} } = \sqrt{106} \: units$

$\small\tt\green{BC}= \sqrt{( - 2 - 3)} {}^{2} + \sqrt{(5 + 4) {}^{2} } = \sqrt{106} \: units$

$\small\tt\red{CA} = \sqrt{(3 - 7)} {}^{2} + \sqrt{( - 4 - 10)} {}^{2} = \sqrt{212 \: units}$

$\small\tt\pink{Clearly,} \\ </p><p>∴ \sqrt{(212)} {}^{2} = \sqrt{(106)} {}^{2} + \sqrt{(106)} {}^{2}$

⇒AC² = AB² + BC²

⇒∠ABC = 90°

‏‏‎ ‎[Converse of Pythagoras theorem]

Here, AB = BC and ∠ABC = 90°

$\small\sf\fbox{So, ΔABC is an isosceles right angle.}$

Hence, Proved

$...$

Answer 2

ANSWER:-

Let A(7, 10), B(-2, 5), C(3, -4) be vertices of an isoceles triangle.

\small\tt\pink{Then \: AB}= \sqrt{(7 + 2)} {}^{2} + \sqrt{(10 - 5) {}^{2} } = \sqrt{106} \: unitsThenAB=(7+2)2+(10−5)2=106units

\small\tt\green{BC}= \sqrt{( - 2 - 3)} {}^{2} + \sqrt{(5 + 4) {}^{2} } = \sqrt{106} \: unitsBC=(−2−3)2+(5+4)2=106units

\small\tt\red{CA} = \sqrt{(3 - 7)} {}^{2} + \sqrt{( - 4 - 10)} {}^{2} = \sqrt{212 \: units}CA=(3−7)2+(−4−10)2=212units

\begin{gathered}\small\tt\pink{Clearly,} \\ < /p > < p > ∴ \sqrt{(212)} {}^{2} = \sqrt{(106)} {}^{2} + \sqrt{(106)} {}^{2}\end{gathered}Clearly,</p><p>∴(212)2=(106)2+(106)2

⇒AC² = AB² + BC²

⇒∠ABC = 90°

‏‏‎ ‎[Converse of Pythagoras theorem]

Here, AB = BC and ∠ABC = 90°

\small\sf\fbox{So, ΔABC is an isosceles right angle.}So, ΔABC is an isosceles right angle.

Hence, Proved