Question

If two vertices of an isosceles right triangle are

A(–1, –7) and B(–7, –1), then find coordinates of

third vertex of the triangle ABC [Given ∠C = 90°].​

Answer 1

Given :   two vertices of an isosceles right triangle ABC are  A(–1, –7) and B(–7, –1),  ∠C = 90°

To Find : coordinates of  C

Solution:

ACB is a right angle triangle at C  and  isosceles triangle

Hence AC = BC

A = ( - 1, -7)

B = (-7 , - 1)

Let say C = ( x, y)

AC & CB are perpendicular to each other

Hence slope of AC * Slope of BC = - 1

=> ( y + 7)/(x + 1)   *  ( y + 1)/(x + 7) = - 1

=> y² + 8y + 7 =  -x² - 8x - 7

=> x² + y²  + 8x + 8y  + 14 = 0

AC² = (x + 1)² + (y + 7)²

BC² = (x + 7)² + (y + 1)²

AC = BC

=> AC² = BC²

=> (x + 1)² + (y + 7)² = (x + 7)² + (y + 1)²

=> x² + 2x + 1 + y² + 49 + 14y  = x² + 14x + 49 + y² + 1 + 2y

=> 12y = 12x

=> x = y

x² + y²  + 8x + 8y  + 14 = 0

=> x² + x²  + 8x + 8x  + 14 = 0

=> x² + 8x + 7 = 0

=> (x + 7)(x + 1) = 0

=> x = - 7 , - 1

=> y = -7 ,   - 1

Hence C  is ( -7 , -7)  or  (-1, - 1)

coordinates of  third vertex of the triangle   ( -7 , -7)  or  (-1, - 1)

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Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\sf{1. \: Distance \: between \: two \: points \: A( x, y) \: and \: (h, k) \: is}$

$= \sqrt{ {(x - h)}^{2} + {(y - k)}^{2} }$

2. Pythagoras theorem states that :

In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides

GIVEN

• Two vertices of an isosceles right triangle ABC are A(–1, –7) and B(–7, –1)
• In the triangle ABC , ∠C = 90°

TO DETERMINE

The coordinates of third vertex of the triangle ABC

CALCULATION

Let ( x, y) be the coordinates of third vertex of the triangle ABC

Here ABC is an isosceles right triangle with ∠C = 90°

So AC = BC

$\implies \: \sqrt{ {(x + 1)}^{2} + {(y + 7)}^{2} } = \sqrt{ {(x + 7)}^{2} + {(y + 1)}^{2} }$

$\implies \: {(x + 1)}^{2} + {(y + 7)}^{2} = {(x + 7)}^{2} + {(y + 1)}^{2}$

$\implies \: {x}^{2} + 2x + 1 + {y}^{2} + 14y + 49 = {x}^{2} + 14x + 49 + {y}^{2} + 2y + 1$

$\implies 12x = 12y$

$\implies \: \: x = y \: \: ........(1)$

Again using Pythagorean Theorem

$\sf{{(AC)}^{2} + {(BC)}^{2} = {(AB)}^{2} }$

$\implies \: \sf{2{(AC)}^{2} = {(AB)}^{2} } \: \: ........(2)$

Again

$\sf{(AB)}^{2} = {( - 1 + 7)}^{2} + {( - 7 + 1)}^{2}$

$\implies \: \sf{(AB)}^{2} =36 + 36$

$\implies \: \sf{(AB)}^{2} =72 \: \: .....(3)$

From Equation (2) using Equation (3)

$\sf{ \implies \: {(x + 1)}^{2} + {(y + 7)}^{2}= 72}$

$\sf{ \implies \: {(x + 1)}^{2} + {(x + 7)}^{2}= 72} \: \: ( \because \: x = y)$

$\sf{ \implies \:2 {x}^{2} + 16x + 14 = 0 }$

$\sf{ \implies \: {x}^{2} + 8x + 7 = 0 }$

$\sf{ \implies \: {x}^{2} + 7x + x+ 7 = 0 }$

$\sf{ \implies \: x(x+ 7) + 1( x+ 7) = 0 }$

$\sf{ \implies \: (x+ 7) ( x+ 1) = 0 }$

$\sf{Which \: gives \: \: x = - 1, - 7}$

From Equation (1)

$\sf{x = - 1 \: \: gives \: \: y = - 1}$

$\sf{x = - 7 \: \: gives \: \: y = - 7}$

RESULT

$\sf{Hence \: the \: required \: point \: is \: ( - 1, - 1) \: \: or \: \: (-7,-7)}$