Question

The value of k, if the point P(2, 4) is equidistant from the points A(k + 1, 3) and B(3, 2k – 1) can be

• A. 1
• B. -2
• C. 4
• D. 5

Answer 1

The correct option is C 4

Given: P(2, 4) is equidistant from A(k + 1, 3) and B(3, 2k – 1)

Distance between points P and A,

$PA=\sqrt{(k+1-2{)}^2+(3-4{)}^2}$

$=\sqrt{k^2+1-2k+1}$

$=\sqrt{k^2-2k+2}$

Distance between points P and B,

$PB=\sqrt{(3-2{)}^2+(2k-1-4{)}^2}$

$=\sqrt{1+4k^2+25-20k}$

$=\sqrt{4k^2-20k+26}$

Given that PA = PB.

$\therefore \sqrt{k^2-2k+2}=\sqrt{4k^2-20k+26}$

$\Rightarrow k^2-2k+2=4k^2-20k+26$

$\Rightarrow 3k^2-18k+24=0$

$\Rightarrow k^2-6k+8=0$

$\Rightarrow k^2-4k-2k+8=0$

$\Rightarrow k(k-4)-2(k-4)=0$

$\Rightarrow (k-2)(k-4)=0$

$\Rightarrow k=2or4$

Hence, the correct answer is option (c).