Question

find the value of k, if the point p (2,4) is equidistant from the point (5,k) and (k,7)​

Answer 1

★Given:-

• The point P(2,4) is equidistant from the points (5,k) and (k,7)​

★To find:-

• Value of k.

★Solution:-

Given, P(2,4)

Let, A(5,k) and B(k,7)​

As point P is equidistant from the points A&B,

  PA = PB

We know:

$\leadsto \underline{\boxed{\sf Distance =\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2} }}$

Now,

$:\implies \sf PA=PB\\\\:\implies \sf \sqrt{(5 - 2)^2 + (k -4)^2} = \sqrt{(k- 2)^2 + (7 -4)^2}$

Squaring on both sides,

$:\implies \sf (5 - 2)^2 + (k -4)^2 = (k- 2)^2 + (7 -4)^2 \\\\:\implies \sf 9 + k^2 + 16- 8k = k^2 + 4 -4k + 9\\\\:\implies \sf k^2- 8k + 16- k^2 -4 + 4k = 0\\\\:\implies \sf -4k + 12 = 0\\\\:\implies \sf 4k = 12\\\\:\implies \sf k=\frac{12}{4} \\\\:\implies \boxed{\boxed{\sf k=3}}$

Hence,

The value of k is 3.

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Answer 2

★Given:-

The point P(2,4) is equidistant from the points (5,k) and (k,7)

★To find:-

Value of k.

★Solution:-

Given, P(2,4)

Let, A(5,k) and B(k,7)

As point P is equidistant from the points A&B,

  PA = PB

We know:

$\leadsto \underline{\boxed{\sf Distance =\sqrt{(x_2-x_1)^2 +(y_2-y_1)^2} }}$

Now,

$:\implies \sf PA=PB\\\\:\implies \sf \sqrt{(5 - 2)^2 + (k -4)^2} = \sqrt{(k- 2)^2 + (7 -4)^2}$

Squaring on both sides,

$:\implies \sf (5 - 2)^2 + (k -4)^2 = (k- 2)^2 + (7 -4)^2 \\\\:\implies \sf 9 + k^2 + 16- 8k = k^2 + 4 -4k + 9\\\\:\implies \sf k^2- 8k + 16- k^2 -4 + 4k = 0\\\\:\implies \sf -4k + 12 = 0\\\\:\implies \sf 4k = 12\\\\:\implies \sf k=\frac{12}{4} \\\\:\implies \boxed{\boxed{\sf k=3}}$

Hence,

The value of k is 3.

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