Question

If the point P(3, 4) is equidistant from the points A(-2,3) and B(k, -1), find the values of k. Also;
find the distance AB.​

Answer 1

Answer:

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Answer 2

The value of k is equa to 4 or, 2.

The distance AB is 4$\sqrt{2}$ units or, $2\sqrt{13}$ units.

Step-by-step explanation:

Given,

The point P(3, 4) is equidistant from the points A(- 2, 3) and B(k, - 1).

To find, the value of k = ? and the distance AB = ?

∴ PA = PB

Using distance formula,

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

PA = $\sqrt{(3+2)^2+(4-3)^2}$

= $\sqrt{(5)^2+(1)^2}$

= $\sqrt{26}$

PB = $\sqrt{(k-3)^2+(-1-4)^2}$

= $\sqrt{k^2-6k+9+25}$

= $\sqrt{k^2-6k+34}$

$\sqrt{k^2-6k+34}$ = $\sqrt{26}$

Squaring both sides, we get

$k^2$ - 6k + 34 = 26

⇒ $k^2$ - 6k + 8 = 0

⇒ $k^2$ - 4k - 2k + 8 = 0

⇒ k(k - 4) -2(k - 4)

⇒ (k - 4)(k - 2) = 0

⇒ k = 4 or, 2

∴ The distance AB =

Using distance formula,

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

Put k = 2

= $\sqrt{(2+2)^2+(-1-3)^2}$ = $\sqrt{16+16}$

= $\sqrt{32}$ = 4$\sqrt{2}$ units

Put k = 4

= $\sqrt{(4+2)^2+(-1-3)^2}$

= $\sqrt{(6)^2+(-4)^2}$

= $\sqrt{36+16}=\sqrt{52}$

= $2\sqrt{13}$ units

Thus, The value of k is equa to 4 or, 2.

The distance AB is 4$\sqrt{2}$ units or, $2\sqrt{13}$ units.