Question

Find the value of k , for which the distance is 5 units for P ( 1, 2) Q ( -2, k )​

Answer 1

Step-by-step explanation:

Distance

PQ = √(x2-x1)²+(y2-y1)²

5 = √(-2-1)²+(k-2)²

25 = (-3)² + (k-2)²

25 = 9 + (k-2)²

(k-2)² = 16

k-2 = 4 & k-2 = -4

k = 6 & k = -2

Answer 2

$\huge{\underline{\underline{\bf{\red{Solution}}}}}$

Applying distance Formula

Distance from PQ = 5unit

$\implies\sf PQ=\sqrt{(x2-x1)^2+(y2-y1)^2}$

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

$\implies\sf 5=\sqrt{(-3)^2+k^2+4-4k}$

$\implies\sf 5=\sqrt{9+k^2+4-4k}$

$\implies\sf 5=\sqrt{k^2-4k+13}$

Squaring both side

$\implies\sf 25=k^2-4k+13$

$\implies\sf k^2-4k-12=0$

$\implies\sf k^2-6k+2k-12=0$

$\implies\sf k(k-6)+2(k-6)=0$

$\implies\sf (k-6)(k+2)=0$

Either

k = 6

or

k = -2