Math, asked by nishi9449, 11 months ago

If A(4,3), B(6,-2)and C(a,-3) are the vertices of a triangle right angled at A,find a.​

Answers

Answered by harendrachoubay
3

The value of a is "\frac{7}{2}".

Step-by-step explanation:

Given,

A(4,3), B(6,-2)and C(a,-3) are the vertices of a triangle right angled at A.

AB^{2} = (6 - 4)^{2} + ( - 2 - 3)^{2} ... (1)

= 2^{2} + 5^{2} = 4 + 25 = 29

BC^{2} = (a - 6)^{2} + ( - 3 + 2)^{2}

= a^{2} - 12a + 36 + 1 = a^{2} - 12a + 37

CA^{2} = (a - 4)^{2} + ( - 3 - 3)^{2}

= a^{2} - 8a + 16 + 36

= a^{2} - 8a + 52

Using (1), We get

∴  a^{2} - 8a + 52 = a^{2} - 12a + 37 + 29

a^{2} - 8a + 52 = a^{2} - 12a + 66

⇒ - 8a + 52 = - 12a + 66

⇒ - 8a + 12a = 66 - 52

⇒ 4a = 14

∴ a = \frac{7}{2}

Hence, the value of a is "\frac{7}{2}".

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