Question

ICA
36. If the point P (k - 1,2) is equidistant from the points A (3,k) and B(k, 5), finde
values of k.
[CBSE 2014]

THE QUESTION IS FROM R.D SHARMA CLASS 10

MY ANSWER IS THE VALUE OF K IS COMING 5/3 BUT IN THE ANSWERS IT IS GIVEN 1,5



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Answer 1

Answer:1 & 5

Step-by-step explanation:

P(k-1,2)

A(3,k)

B(k,5)

ATQ,

AP = BP

$\sqrt ((3-k+1)^{2}$ +$(k-2)^{2})$ = $\sqrt((k-k+1)+(5-2)^{2})$

$(4-k)^{2} + (k-2)^{2} = 1+9$

$16-8k+k^{2}+k^{2}-4k+4=10$

$2k^{2}-12k+10=0\\ k^{2}-6k+5=0 \\k^{2}-5k-k+5=0\\ k(k-5)-1(k-5)=0\\(k-5)(k-1)=0\\\\k=5 k=1$