Question

Find the values of $y$ for which the distance between the points P$\left( {2, - 3} \right)$ and Q$\left( {10,y} \right)$ is $10$ Units.
A) $9$
B) $3$
C) $6$
D) None of these

Answer 1

Hint: In the above question two points P and Q are given. For this, we have to find the distance between them and find the value of$y$. Here we use the formula of the distance between two points the calculate the value of$y$. Put the values in the formula of distance and get the correct answer.
Formula: $d = {\sqrt {\left( {{x_2} - {x_1}} \right)^2 + {\left( {{y_2} - {y_1}} \right)^2}}}$

Complete step by step solution: Given that: P$\left( {2, - 3} \right)$ and Q$\left( {10,y} \right)$
And given PQ$ = 10$ Units
Where
 $
  {x_1} = 2 \\
  {x_2} = 10 \\
  {y_1} = \left( { - 3} \right) \\
  {y_2} = y \\
 $
By using the formula of distance between two points:
$d = {\sqrt {\left( {{x_2} - {x_1}} \right)^2 + {\left( {{y_2} - {y_1}} \right)^2}}}$
Where
$d = 10$
$d = {\sqrt {\left( {10-2} \right)^2 + {\left( {{y} - \left( {-3} \right )} \right)^2}}}$
Squaring on both sides of equation:
We get:
$
  100 = {\left( 8 \right)^2} + {\left( {y + 3} \right)^2} \\
   \Rightarrow 100 = 64 + {\left( {y + 3} \right)^2} \\
   \Rightarrow 100 - 64 = {\left( {y + 3} \right)^2} \\
   \Rightarrow {\left( {y + 3} \right)^2} = 36 \\
   \Rightarrow \left( {y + 3} \right) = \pm 6 \\
 $
It means
$
  y + 3 = 6 \\
  {\text{or}} \\
  y + 3 = - 6 \\
    \\
 $
From the above statements we get:
If
$
  y + 3 = 6 \\
  y = 6 - 3 \\
   \Rightarrow y = 3 \\
 $
And If
$
  y + 3 = - 6 \\
  y = - 6 - 3 \\
   \Rightarrow y = \left( { - 9} \right) \\
 $
Hence we get two values of $y$
$
  y = 3 \\
  {\text{and}} \\
  y = \left( { - 9} \right) \\
 $

Hence option D is the correct answer.

Note: First we have to remember the formula of distance between two points. That is $d = {\sqrt {\left( {{x_2} - {x_1}} \right)^2 + {\left( {{y_2} - {y_1}} \right)^2}}}$. Then from the given points P and Q write the values of${x_{1,}}{x_{2,}}{y_1}\& {y_2}$. Then put these values in the formula of distance and find the value of$y$. Thus we get the two value of$y$. One is positive value and another is negative value.