Question

Point A (a, 6) and B (2,8) are equidistant from the point C (1,1). Find a.​

Answer 1

The value of a is 6 or - 4.

Step-by-step explanation:

To find, the value of a = ?

The point A (a, 6) and B (2, 8) are equidistant from the point C (1, 1).

∴ $PA^{2} =PB^{2}$

We know distance formula,

$\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

$(1-a)^{2}+(1-6)^{2} =(1-2)^{2}+(1-8)^{2}$

⇒ $(1-a)^{2}+(-5)^{2} =(-1)^{2}+(-7)^{2}$

⇒$a^{2} -2a+1 +25 =1+49$

⇒$a^{2} -2a-24=0$

⇒ $a^{2} -6a+4a-24=$

⇒ $(a-6)(a+4)=0$

∴ a = 6 or - 4

Hence, the value of a is 6 or - 4.