Question

(b)
Given three points P(-1,2), A (2, k) and B(k-1). Given the PA = PB.
Find the value of k.​

Answer 1

Answer

$\sf\frac{1}{2}$

Explanation

Given,

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

PA = PB

To find,

The value of k

Apply Distance formula,

$\sf \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Then,

for PA, distance = $\sf\sqrt{(2 - (-1))^2 + (k - 2)^2}$

and, for PB, distance = $\sf\sqrt{(k - (-1))^2 + (-1 - 2)^2}$

Equating the two,

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

Square both sides

→ $\sf (3)^2 + (k - 2)^2 = (k + 1)^2 + (-3)^2$

→ 9 + k² + 4 - 4k = k² + 1 + 2k + 9

→ 4 - 4k = 1 + 2k

→ 4 - 1 = 4k + 2k

→ 6k = 3

→ k = ½

Answer 2

Answer:

$\large\boxed{\sf{k=\dfrac{1}{2}}}$

Step-by-step explanation:

Given three points are :-

P ( -1,2 )

A ( 2, k )

B (k, -1 )

Also, it's given that, PA = PB

By distance formula, we have,

$= > PA = \sqrt{ {( - 1 - 2)}^{2} + {(2 - k)}^{2} } \\ \\ = > PA = \sqrt{ {( - 3)}^{2} + {(2 - k)}^{2} } \\ \\ = > PA = \sqrt{9 + {(2 - k)}^{2} } \: \: \: \: \: \: \: .........(1)$

Similarly, we have,

$= > PB = \sqrt{ {( - 1 - k)}^{2} + {(2 + 1)}^{2} } \\ \\ = > PB = \sqrt{ {( - 1 - k)}^{2} + {(3)}^{2} } \\ \\ = > PB = \sqrt{ {( - 1 - k)}^{2} + 9} \: \: \: \: \: \: \: ........(2)$

°.° PA = PB

.°. from Equation (1) and (2) , we have

$= > \sqrt{ 9 + {(2 - k)}^{2} } = \sqrt{ {( - 1 - k)}^{2} + 9} \\ \\ = > 9 + {(2 - k)}^{2} = 9 + {( - 1 - k)}^{2} \\ \\ = > {(2 - k)}^{2} = {( - 1 - k)}^{2} \\ \\ = > 4 + {k}^{2} - 4k = 1 + {k}^{2} + 2k \\ \\ = > 4k + 2k = 4 - 1 \\ \\ = > 6k = 3 \\ \\ = > k = \frac{3}{6} \\ \\ = > k = \frac{1}{2}$

Concept Map :-

• Distance between two points ( a , b ) and ( c , d ) is given by, $\sf{ \sqrt{ {(a - c)}^{2} + {(b - d)}^{2} } }$