Question

Heyya ❤

Just a simple question :

☞ Show that the points A ( 7 , 10 ) , B ( -2 , 5 ) , C ( 3, -4 ) are the vertices of a right angled triangle.

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@ llAbdulkadarll​

Answer 1

Given :

• A = $( 7 , 10 ) , B = ( -2 , 5 )\:and\:C = ( 3 , -4 )$

According to the question :

Here A = $( 7 , 10 ) , B = ( -2 , 5 )\:and\:C = ( 3 , -4 )$

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AB = $\sqrt {( x_2 - x_1 )² + ( y_2 - y_1 )²}$

⟹ $\sqrt {( - 2 - 7 )² + ( 5 - 10 )²}$

⟹ $\sqrt {( -9 )² + ( -5 )²}$

⟹ $\sqrt {( 81 + 25 )}$

⟹ $\sqrt {( 106 )}$

AB² = 106 ...... [ 1 st equation ]

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BC = $\sqrt {( x_2 - x_1 )² + ( y_2 - y_1 )²}$

⟹ $\sqrt {( 3 - ( -2 ) )² + ( - 4 - 5 )²}$

⟹ $\sqrt {( 5 )² + ( - 9 )²}$

⟹ $\sqrt {25+ 81}$

⟹ $\sqrt {106}$

BC² = 106 ...... [ 2nd Equation ]

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AC = $\sqrt {( x_2 - x_1 )² + ( y_2 - y_1 )²}$

⟹ $\sqrt {( 3 - 7 )² + ( - 4 - 10)²}$

⟹ $\sqrt {( - 4 )² + ( - 14 )²}$

⟹ $\sqrt {16 + 196 }$

⟹ $\sqrt {212}$

AC² = 212 ...... [ 3rd Equation ]

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From [ 1 , 2 and 3 equations ] We get :

➳ AB² + BC²

➳ 106 + 106

➳ $212 = AC²$

Since, AB² + BC² = AC²

∴ ΔABC is a right angled triangle, right angled at B.

So, It's Done !!

Answer 2

$\underline{\bf{ \: Given:-}}$

• Points A(7,10), B(-2,5) and C(3,-4)

$\underline{\bf{ \: To \: Find:-}}$

• Given points are the vertices of right angled Triangle.

$\underline{\bf{ \: Solution:-}}$

Here,

we, know that

$\large{\boxed{\sf Distance \: Formula = \sqrt{ {(x_{2} - x_{1})}^{2} + {( y_{2} - y_{1} ) }^{2} } \: }}$

So,

$\ast\sf AB = \sqrt{ {(x_{2} - x_{1})}^{2} + {( y_{2} - y_{1} ) }^{2} }$

where,

• $\sf x_1 = 7$
• $\sf x_2 = -2$
• $\sf y_1 = 10$
• $\sf y_2 = 5$

So,

$\dashrightarrow\sf AB = \sqrt{ {( - 2 -7)}^{2} + {( 5 - 10) }^{2} }$

$\dashrightarrow\sf AB = \sqrt{ {( -9)}^{2} + {(- 5) }^{2} }$

$\dashrightarrow\sf AB = \sqrt{(81)+ (25)}$

$\dashrightarrow\sf AB = \sqrt{81+ 25}$

$\dashrightarrow\sf AB = \sqrt{106}$

$\dashrightarrow\sf {AB}^{2} = 106 \huge{......1}$

_____________________________

Now,

$\ast\sf BC = \sqrt{ {(x_{2} - x_{1})}^{2} + {( y_{2} - y_{1} ) }^{2} }$

where,

• $\sf x_1 = -2$
• $\sf x_2 = 3$
• $\sf y_1 = 5$
• $\sf y_2 = -4$

So,

$\dashrightarrow\sf BC = \sqrt{ {(3 -( - 2))}^{2} + {( - 4 - 5) }^{2} }$

$\dashrightarrow\sf BC = \sqrt{ {(3 + 2)}^{2} + {( - 4 - 5) }^{2} }$

$\dashrightarrow\sf BC = \sqrt{ {(5)}^{2} + {( -9) }^{2} }$

$\dashrightarrow\sf BC = \sqrt{25 + 81 }$

$\dashrightarrow\sf BC = \sqrt{106}$

$\dashrightarrow\sf {BC}^{2} = 106 \huge{.......2}$

_____________________________

$\ast\sf AC = \sqrt{ {(x_{2} - x_{1})}^{2} + {( y_{2} - y_{1} ) }^{2} }$

where,

• $\sf x_1 = 7$
• $\sf x_2 = 3$
• $\sf y_1 = 10$
• $\sf y_2 = -4$

So,

$\dashrightarrow\sf AC = \sqrt{ {(3- 7)}^{2} + {( - 4 - 10) }^{2} }$

$\dashrightarrow\sf AC = \sqrt{ {( - 4)}^{2} + {( -14) }^{2} }$

$\dashrightarrow\sf AC = \sqrt{16 + 196}$

$\dashrightarrow\sf AC = \sqrt{212}$

$\dashrightarrow\sf {AC}^{2} =212 \huge{.......3}$

_____________________________

we, know

$\huge{ \boxed{\bf{H^2 = P^2 + B^2}}}$

So,

$:\to\large{\sf {AC}^2 = {AB}^2 + {BC}^2}$

where,

• AC² = 212
• AB² = 106
• BC² = 106

[FROM EQ. 1,2 AND 3]

$:\to\large{\sf {AC}^2 = {AB}^2 + {BC}^2}$

$:\to\large{\sf 212= 106 + 106}$

$:\to\large{\sf 212= 212}$

Here,

Square of Hypotenuse is equal to the sum of square of other two sides.

Hence, Given Vertices are points of right angled Triangle.