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Question: Find the values of \(y\) for which the distance between the points P\(\left( {2, - 3} \right)\) and ...

Find the values of yy for which the distance between the points P(2,−3)\left( {2, - 3} \right) and Q(10,y)\left( {10,y} \right) is 1010 Units.
A) 99
B) 33
C) 66
D) None of these

Explanation

Solution

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 ofyy. Here we use the formula of the distance between two points the calculate the value ofyy. Put the values in the formula of distance and get the correct answer.
Formula: d=(x2−x1)2+(y2−y1)2d = {\sqrt {\left( {{x_2} - {x_1}} \right)^2 + {\left( {{y_2} - {y_1}} \right)^2}}}

Complete step by step solution: Given that: P(2,−3)\left( {2, - 3} \right) and Q(10,y)\left( {10,y} \right)
And given PQ=10 = 10 Units
Where
x1=2 x2=10 y1=(−3) y2=y  {x_1} = 2 \\\ {x_2} = 10 \\\ {y_1} = \left( { - 3} \right) \\\ {y_2} = y \\\
By using the formula of distance between two points:
d=(x2−x1)2+(y2−y1)2d = {\sqrt {\left( {{x_2} - {x_1}} \right)^2 + {\left( {{y_2} - {y_1}} \right)^2}}}
Where
d=10d = 10
d=(10−2)2+(y−(−3))2d = {\sqrt {\left( {10-2} \right)^2 + {\left( {{y} - \left( {-3} \right )} \right)^2}}}
Squaring on both sides of equation:
We get:
100=(8)2+(y+3)2 ⇒100=64+(y+3)2 ⇒100−64=(y+3)2 ⇒(y+3)2=36 ⇒(y+3)=±6  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 or y+3=−6   y + 3 = 6 \\\ {\text{or}} \\\ y + 3 = - 6 \\\ \\\
From the above statements we get:
If
y+3=6 y=6−3 ⇒y=3  y + 3 = 6 \\\ y = 6 - 3 \\\ \Rightarrow y = 3 \\\
And If
y+3=−6 y=−6−3 ⇒y=(−9)  y + 3 = - 6 \\\ y = - 6 - 3 \\\ \Rightarrow y = \left( { - 9} \right) \\\
Hence we get two values of yy
y=3 and y=(−9)  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=(x2−x1)2+(y2−y1)2d = {\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 ofx1,x2,y1&y2{x_{1,}}{x_{2,}}{y_1}\& {y_2}. Then put these values in the formula of distance and find the value ofyy. Thus we get the two value ofyy. One is positive value and another is negative value.