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Question: Find the value of y, if the distance between the points (2, y) and (−4, 3) is 10 units....

Find the value of y, if the distance between the points (2, y) and (−4, 3) is 10 units.

Explanation

Solution

Hint : We have been given the coordinates of the two points. The distance between these points is calculated using the distance formula. By substituting all the known in this formula, we can calculate the unknown one.
Formula to be used:
d=(x2−x1)2+(y2−y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} where, d is the distance between the two points and x and y represent the respective coordinates of both the points.

Complete step-by-step answer :
Let the two given points be A and B, their coordinates are:
A (2, y)
B (−4, 3)
The distance between these points can be calculated using the distance formula.
For the two points with coordinates (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) , the distance (d) between them is calculated by using the distance formula given as:
d=(x2−x1)2+(y2−y1)2d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}
Here, for the points A and B, the respective coordinates are:
x1=2 x2=−4   {x_1} = 2 \\\ {x_2} = - 4 \;
y1=y y2=3   {y_1} = y \\\ {y_2} = 3 \;
The distance between them is given to be 10 units.
d = 10
Substituting these values in the distance formula, we get:
⇒10=(−4−2)2+(3−y)2 ⇒10=(−6)2+(3−y)2 ⇒10=36+(3−y)2  \Rightarrow 10 = \sqrt {{{\left( { - 4 - 2} \right)}^2} + {{\left( {3 - y} \right)}^2}} \\\ \Rightarrow 10 = \sqrt {{{\left( { - 6} \right)}^2} + {{\left( {3 - y} \right)}^2}} \\\ \Rightarrow 10 = \sqrt {36 + {{\left( {3 - y} \right)}^2}} \\\
Squaring both the sides for simplification and calculating the value of y:
⇒100=36+(3−y)2 ⇒(3−y)2=100−36 ⇒(3−y)2=64 ⇒(3−y)2=(8)2 ⇒3−y=±8 ⇒y=3∓8  \Rightarrow 100 = 36 + {\left( {3 - y} \right)^2} \\\ \Rightarrow {\left( {3 - y} \right)^2} = 100 - 36 \\\ \Rightarrow {\left( {3 - y} \right)^2} = 64 \\\ \Rightarrow \sqrt {{{\left( {3 - y} \right)}^2}} = \sqrt {{{\left( 8 \right)}^2}} \\\ \Rightarrow 3 - y = \pm 8 \\\ \Rightarrow y = 3 \mp 8 \\\
This shows that there are two possible values of y:
i)y=3−8 ⇒y=−5 ii)y=3+8 ⇒y=11   i)y = 3 - 8 \\\ \Rightarrow y = - 5 \\\ ii)y = 3 + 8 \\\ \Rightarrow y = 11 \;
Therefore if the distance between the points (2, y) and (−4, 3) is 10 units, the value of y can be both – 5 and 11.
So, the correct answer is “ – 5 and 11.”.

Note : As we have to find the distance between A and B, we took A as the first point and its coordinates were x1,y1{x_1},{y_1} . But, if t was given in opposite order, B and A, then B would have been treated as the first point.
Whenever we take the square root of a squared quantity, it is always taken with a negative - positive sign because squares of negative numbers are also positive.
(8)2=(−8)2=64 ⇒64=±8   {\left( 8 \right)^2} = {\left( { - 8} \right)^2} = 64 \\\ \Rightarrow \sqrt {64} = \pm 8 \;
When the sign ′±′' \pm ' goes from one to another side of the equation, it becomes ′∓′' \mp '