Question

Given 3 points P(-1, 2), A(2, k) and B(k, -1). PA=PB.Find K

Answer 1

Answer:

Step-by-step explanation:

Here is your answer.

Answer 2

Question :-

Given three points P(-1,2) , A(2,k) and B(k ,-1)

if PA = PB ,then find the value of k .

Answer :-

$\to \boxed{ \sf k = \frac{1}{2} } \:$

To Find :-

→ Value of k .

Used formula :-

$\sf if \: given \: two \: pionts \:A( x_1 , y_1) \: and \:B (x_2 , y_2) \\ \\ \sf distance \: between \:AB \: is \: - \\ \\ \to \sf \: AB = \sqrt{ {(x_2 - x_1)}^{2} + {(y_2 - y_1)}^{2} }$

Solution :-

According to the question

→ P(-1,2) , A(2,k) and B(k ,-1)

Firstly we will have to find distance between P and A -

$\to \sf P( - 1 , 2) \mapsto \: x_1 = - 1 , y_1 = 2 \\ \\ \to \sf A( 2 , k)\mapsto \: \: x_2 = 2 , y_2= k \: \\ \\ \to \sf \: PA = \sqrt{ {(2 + 1)}^{2} + {(k - 2)}^{2} } \: .....eq.(1)\\ \\ \bf \: and \: \\ \\ \to \sf P( - 1 , 2) \mapsto \: x_1 = - 1 , y_1 = 2 \: \\ \\ \to \sf B ( k , - 1)\mapsto \: \: x_2 = k , y_2= - 1 \: \\ \\ \to \sf \: PB \: = \sqrt{ {(k + 1)}^{2} + {( - 1 - 2)}^{2} } \: ... eq.(2)$

According to the question

→ PA = PB

hence ,from eq.(1) and (2)

$\sf \sqrt{ {(2 + 1)}^{2} + {(k - 2)}^{2} } = \sqrt{ {(k + 1)}^{2} + {( - 1 - 2)}^{2} } \: \\ \\ \sf squaring \: on \: both \: sides \\ \\ \to \sf {3}^{2} + {(k - 2)}^{2} = {(k + 1)}^{2} + {( - 3)}^{2} \\ \\ \to \sf \: 9 + {k}^{2} + 4 - 4k = {k}^{2} + 1 + 2k + 9 \\ \\ \to \sf 13 - 4k - 2k = 10 \\ \\ \to \sf - 6k = 10 - 13 \\ \\ \to \sf - 6k = - 3 \\ \\ \to \boxed{ \sf k = \frac{1}{2} }$