If the point $P (k – 1, 2)$ is equidistant from the points $A (3, k)$ and $B (k, 5)$, find the values of $k$.
Given:
The point $P (k-1, 2)$ is equidistant from the points $A(3, k)$ and $B(k, 5)$.
To do:
We have to find the value of $k$.
Solution:
$PA$ is equidistant from $PB$.
This implies,
$PA=PB$
Squaring on both sides, we get,
$PA^2=PB^2$
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( \mathrm{PA}=\sqrt{(k-1-3)^{2}+(2-k)^{2}} \)
Squaring on both sides, we get,
\( \mathrm{PA}^{2}=(k-4)^{2}+(2-k)^{2} \)
\( =k^2-8k+16+4+k^{2}-4 k \)
\( =2k^{2}-12k+20 \)
\( \mathrm{PB}^{2}=(k-1-k)^{2}+(2-5)^{2} \)
\( =(-1)^{2}+(-3)^{2} \)
\( =1+9 \)
\( =10 \)
\( \Rightarrow 2k^{2}-12 k+20=10 \)
\( \Rightarrow 2(k^2-6k+10)=2(5) \)
\( \Rightarrow k^2-6k+10-5=0 \)
\( k^2-6k+5=0 \)
\( k^2-k-5k+5=0 \)
\( k(k-1)-5(k-1)=0 \)
\( (k-5)(k-1) =0 \)
\( k=5 \) or \( k=1 \)
The values of $k$ are $1$ and $5$. 
Related Articles
- Find the value of $k$, if the point $P (0, 2)$ is equidistant from $(3, k)$ and $(k, 5)$.
- If the point $P (2, 2)$ is equidistant from the points $A (-2, k)$ and $B (-2k, -3)$, find $k$. Also, find the length of AP.
- Find the values of \( k \) if the points \( \mathrm{A}(k+1,2 k), \mathrm{B}(3 k, 2 k+3) \) and \( \mathrm{C}(5 k-1,5 k) \) are collinear.
- Find the value (s) of $k$ for which the points $(3k – 1, k – 2), (k, k – 7)$ and $(k – 1, -k – 2)$ are collinear.
- Find the value of $k$, if the points $A( 8,\ 1),\ B( 3,\ -4)$ and $C( 2,\ k)$ are collinear.
- Find the values of $k$, if the points $A( k+1,\ 2k),\ B( 3k,\ 2k+3)$ and $C( 5k+1,\ 5k)$ are collinear.
- The line joining the points $(2, 1)$ and $(5, -8)$ is trisected at the points P and Q. If point P lies on the line $2x – y + k = 0$. Find the value of $k$.
- Find the value of $k$ if points $(k, 3), (6, -2)$ and $(-3, 4)$ are collinear.
- If the points $( \frac{2}{5},\ \frac{1}{3}),\ ( \frac{1}{2},\ k)$ and $( \frac{4}{5},\ 0)$ are collinear, then find the value of $k$.
- Find the value of \( k \), if \( x-1 \) is a factor of \( p(x) \) in each of the following cases:(i) \( p(x)=x^{2}+x+k \)(ii) \( p(x)=2 x^{2}+k x+\sqrt{2} \)(iii) \( p(x)=k x^{2}-\sqrt{2} x+1 \)(iv) \( p(x)=k x^{2}-3 x+k \)
- In each of the following find the value of $k$ for which the points are collinear.(i) $(7, -2), (5, 1), (3, k)$(ii) $(8, 1), (k, -4), (2, -5)$
- Find the value of $k$ if $x - 3$ is a factor of $k^2x^3 - kx^2 + 3kx - k$.
- If the points $A (6, 1), B (8, 2), C (9, 4)$ and $D (k, p)$ are the vertices of a parallelogram taken in order, then find the values of $k$ and $p$.
- Find the value of $k$, if the points $A (7, -2), B (5, 1)$ and $C (3, 2k)$ are collinear.
- The point A divides the join of $P (-5, 1)$ and $Q (3, 5)$ in the ratio $k : 1$. Find the two values of $k$ for which the area of $\triangle ABC$ where $B$ is $(1, 5)$ and $C (7, -2)$ is equal to 2 units.
Kickstart Your Career
Get certified by completing the course
Get Started