Question

If the point P (k -1,2) is equidistant from the point A (3,k ) and B(k,5), find the value of k

Answer 1

$\huge\sf\pink{Answer}$

☞ Value of k is 5 or 1

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$\huge\sf\blue{Given}$

✭ P(1,2) is equidistant from A(3,k) & B(k,5)

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$\huge\sf\gray{To \:Find}$

◈ The Value of k?

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$\huge\sf\purple{Steps}$

☯ $\underline{\sf As \ Per \ the \ Question}$

≫ PA = PB

Squaring both sides,

➳ $\sf(PA)^2 \: = (PB)^2$

Let's use distance formula, that is,

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

Distance of PA

≫ $\sf{PA = \sqrt{ (3 -(k -1))^2 + ( k-2)^2}}$

Distance of PB

➳ $\sf PB\: = \: \sqrt{ (k-(k-1))^2 +(5-2)^2}$

On Squaring both sides

➳ $\sf (PA)^2 \: = \: (PB)^2$

➳ $\sf\sqrt{\bigg\lgroup(3 -(k-1))^2 +(k-2)^2\bigg\rgroup^2}$

➳ $\sf\sqrt{\bigg\lgroup k-(k-1)^2 +(5-2)^2\bigg\rgroup^2}$

➳ $\sf{(3-(k-1))^2 +(k-2)^2}$

➳ $\sf (k-(k-1)^2 +(5-2)^2$

➳ $\sf (3-k+1)^2 + (k-2)^2$

➳ $\sf (k-k+1)^2 - (3)^2$

➳ $\sf (4-k)^2 + (k-2)^2 = 10$

$\bigg\lgroup\sf (a-b)^2 = z^2 - 2ab + b^2\bigg\rgroup$

➳ $\sf 16 + k^2 - 8k + k^2 + 4 - 4k = 10$

➳ $\sf 2(k^2 - 6k + 5) \: = \: 0$

➳ $\sf k^2 - 6k + 5 = 0$

Using splitting middle term

➝ $\sf k^2 - 5k -k + 5 = 0$

➝ $\sf k(k-5) - 1(k-5) = 0$

➝ $\sf (k-5)(k-1) = 0$

➝ $\sf\red{k = 5}$

➝ $\sf\orange{k = 1}$

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Answer 2

$\huge\underline{\overline{\mid{\bold{\red{ANSWER-}}\mid}}}$

Given, P(0, 2) is equidistant from A(3, k) and B(k, 5).

∴ AP = PB

⇒ AP^2 = PB^2

⇒ (3 – 0)^2 + (k – 2)^2 = (k – 0)^2 + (5 – 2)^2 [ Using Distance formula]

⇒ 9 + k^2 – 4k + 4 = k^2 + 9

⇒ – 4k + 4 = 0

⇒ 4k = 4

⇒ k = 1

Thus, the value of k is 1.