Question

A(1,7), B(2,4), C(k,5) are the vertices of a right angled ΔABC,Find k, if ∠C is a right angle

Answer 1

Dear Student,

Solution:

If given points are vertex of right angle triangle,then we can apply Pythagoras theorem

AB² = AC²+BC²

now you have vertex of all three points, apply distance formula to calculate the distance of AB, AC and BC

AB = √{(2-1)² + (4-7)²} = √10

AB² = 10 -------eq1

BC = √{(2-k)² + (4-5)²} = √{(2-k)² +1}

BC² = (2-k)²+1 ----------eq2

AC = √{(1-k)² + (7-5)²} = √{(1-k)² +4}

AC² = (1-k)² +4------------eq3

Apply all distances into Pythagoras theorem

AB² = AC²+BC²

10 = (1-k)² +4 +(2-k)²+1

10 = 1 +k² -2k +4+ 4 +k²-4k+1

- 2k² +6k = 0

2k( -k+3) =0

2k = 0

k = 0

or -k+3 =0

-k = -3

k = 3

The point can be ( 0,5) and ( 3,5)

hope it helps you.

Answer 2

____________________________

In ∆ABC , <C = 90°

We know that ,

Let the two points A( x1, y1 ) , B(x2, y2 )

distance between then AB .

AB = √( x2 - x1 )² + ( y2 - y1 )²

______________________________

Now ,

i ) A( 1 , 7 ), B( 2 , 4 )

AB = √( 2 - 1 )² + ( 4 - 7 )²

= √ 1² + ( -3 )²

= √ 1 + 9

= √10

AB² = 10 ---( 1 )

ii ) A( 1 , 7 ), C( k , 5 )

AC = √( k - 1 )² + ( 5 - 7 )²

= √ k² + 1 - 2k + 4

= √ k² - 2k + 5

AC² = k² - 2k + 5 ----( 2 )

iii ) C( k , 5 ) , B( 2 , 4 )

CB = √ ( 2 - k )² + ( 4 - 5 )²

= √ 2² + k² - 4k + 1

= √ k² - 4k + 5

CB² = k² - 4k + 5 ------( 3 )

By Phythogarian theorem ,

AB² = AC² + CB²

From ( 1 ) , ( 2 ) and ( 3 ), we get

10 = k² - 2k + 5 + k² - 4k + 5

=> 10 = 2k² - 6k + 10

=> 0 = 2k² - 6k + 10 - 10

=> 2k² - 6k = 0

=> 2k( k - 3 ) = 0

=> 2k = 0 or k - 3 = 0

=> k = 0 or k = 3

Therefore ,

k = 0 or k = 3

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