Question

Prove that the points A ( - 5,4 ); B ( - 1, - 2 ) and C ( 5,2 ) are the vertices of an isosceles right - angled triangle. Find the co-ordinates of D so that ABCD is a square.

Answer 1

hello friend
we have two equations. subtract both (1) and( 2) we get
x² + y² + 10x - 8y - 11 = 0
x² + y² - 10x - 4y - 23= 0
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20x - 4y +12 = 0
here , x = 4y- 12
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20
put value of x in eq. (1)
( 4y-12/20)² + y² +10( 4y-12/20) - 8y -11= 0
then get the value of y......
hope it helps you.....

Answer 2

concept :- any triangle will be isosceles
right angle triangle when two sides of triangle are equal ,and one angle is right angle . it means all three sides follow Pythagoras theorem .

solution :-
given ,
A(-5, 4) , B(-1 , -2) and C(5 , 2)

using distance formula for finding distance between two points .
AB = √{(-5+1)² + (4+2)²} = √(4²+6²) = 2√13
BC =√{(-1-5)²+(-2-2)²} =√(6²+4²) =2√13
CA =√{5+5)²+(2-4)²} =√(10²+2²) =2√26

now,
here we see that,
AB = BC
it means triangle is an isosceles triangle .
again,
we see that
CA² = 104
AB² + BC² = 104
e.g AB² + BC² = CA²
follow Pythagoras theorem ,
so, triangle is also right angle triangle

hence, ABC is an isosceles right angle ∆.


ABCD is square then,
AB = BC = CD =DA
and also co-ordinate of midpoint of diagonal always concide.
here, AC and BD are diagonals of square
so, midpoint of AC = midpoint of BD
use section formula,
Let D = (x , y)

{(-5 + 5)/2 , (4+2)/2 } = {(x -1)/2 , (y-2)/2}

0 =( x - 1)/2

x = 1
and (y -2)/2 = (4+2)/2 = 6/2

y = 8
hence , D=( 1 ,8)