Question

Find k if A(K, 2), B(3, 1), C(4, 2) are the vertices of a right angle triangle.
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Answer 1

The value of k is equal to "2".

Step-by-step explanation:

Here, A(K, 2), B(3, 1), C(4, 2) are the vertices of a right angle triangle.

$AB^{2} = (3 - K)^{2}$ + (1 - 2)^{2}[/tex]

= (9 - 6K + $K^{2}$) + 1

= $K^{2}$) - 6K + 10

$BC^{2} = (4 - 3)^{2}$ + (2 - 1)^{2}[/tex]

= (1)^{2}[/tex] + (1)^{2}[/tex]

= 1 + 1 =2

$CA^{2} = (4 - K)^{2}$ + (2 - 2)^{2}[/tex]

= 16 - 8K + $K^{2}$

= $K^{2}$ - 8K + 16

∴ In right angled triangleABC,

$CA^{2} = [tex]AB^{2} + [tex]BC^{2}</p><p>⇒ [tex]K^{2}$ - 8K + 16 = ($K^{2}$) - 6K + 10) + 2

⇒  - 8K + 16 = - 6K + 12

⇒  2K = 4

⇒ K = 2

Hence, the value of K is equal to "2".