Question

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

Answer 1

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 "$\frac{7}{2}$".