Question

If a point A (0,2) is equidistant from the points B (3,p) and C (p,5), then find the values of p.

Answer 1

Hint: We will use the distance formula $OP=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ between two coordinates $O\left( {{x}_{1}},{{y}_{1}} \right)$ and $P\left( {{x}_{2}},{{y}_{2}} \right)$ to find the distance between coordinates A and B, and A and C. Then we will equate the values of AB and AC to find the value of p.

Complete step by step answer:
We know the distance between two coordinates $O\left( {{x}_{1}},{{y}_{1}} \right)$ and $P\left( {{x}_{2}},{{y}_{2}} \right)$ on a xy plane can be found by formula
$OP=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}\text{ }\ldots \left( i \right)$
For the coordinates A (0,2) and B (3,p), the distance formula mentioned in formula (i) can be used as
\[\begin{align}
  & AB=\sqrt{{{\left( 3-0 \right)}^{2}}+{{\left( p-2 \right)}^{2}}} \\
 & =\sqrt{{{\left( 3 \right)}^{2}}+{{\left( p-2 \right)}^{2}}} \\
 & =\sqrt{9+{{p}^{2}}-4p+4} \\
 & AB=\sqrt{{{p}^{2}}-4p+13}\text{ }\ldots (ii)
\end{align}\]
For the coordinates A (0,2) and C (p,5), the distance formula mentioned in formula (i) can be used as
$\begin{align}
  & AC=\sqrt{{{\left( p-0 \right)}^{2}}+{{\left( 5-2 \right)}^{2}}} \\
 & =\sqrt{{{\left( p \right)}^{2}}+{{\left( 3 \right)}^{2}}} \\
 & AC=\sqrt{{{p}^{2}}+9}\text{ }\ldots (iii)
\end{align}$
It is given that the point A is equidistant from B and C. This means that the distance AB and AC are the same. This implies that the values of AB and AC found in equations (ii) and (iii) are the same.
Hence, equating equations (ii) and (iii), we get
\[\begin{align}
  & \text{ }AB=AC \\
 & \Rightarrow \sqrt{{{p}^{2}}-4p+13}=\sqrt{{{p}^{2}}+9} \\
\end{align}\]
Squaring these, we get
\[\begin{align}
  & {{\left[ \sqrt{{{p}^{2}}-4p+13} \right]}^{2}}={{\left[ \sqrt{{{p}^{2}}+9} \right]}^{2}} \\
 & \Rightarrow {{p}^{2}}-4p+13={{p}^{2}}+9 \\
\end{align}\]
Solving further, we get
-4p+13=9
-4p=9-13
 -4p=-4
   p=1

Hence, the value of p is 1.

Note: Distance is a non-negative quantity, which means, it is either zero or positive. Thus, while taking square roots, we will always consider positive square roots. If the answer is coming in negative, check for calculation mistakes. If the calculations and formula used is correct, then simply give the modulus on either side of the equation and state in the answer that ‘since distance can’t be negative, we will consider only the numeric value of the answer’.