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Question: If the point \[A(2, - 4)\] is equidistant from \( P(3,8) \) and \( Q( - 10,y) \) . Find the value of...

If the point A(2,−4)A(2, - 4) is equidistant from P(3,8)P(3,8) and Q(−10,y)Q( - 10,y) . Find the value of y. Also find PQ.

Explanation

Solution

Hint : Distance is the measure of how far two points or objects are located. It can also be defined as the length of a straight line joining the two points. It can be measured using instruments like a ruler. But when the coordinates of the two points are given, it can also be calculated using the distance formula. With the help of the distance formula we can find out the value of y and thus PQ.

Complete step-by-step answer :
Distance of between two points is given by the formula,
distance=(x2−x1)2+(y2−y1)2distance = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}
Distance between A(2,−4)A(2, - 4) and P(3,8)P(3,8) is
AP=(3−2)2+(8−(−4))2=(1)2+(12)2=145AP = \sqrt {{{(3 - 2)}^2} + {{(8 - ( - 4))}^2}} = \sqrt {{{(1)}^2} + {{(12)}^2}} = \sqrt {145}
Distance between A(2,−4)A(2, - 4) and Q(−10,y)Q( - 10,y) is
AQ=(−10−2)2+(y−(−4))2=(12)2+(y+4)2=144+y2+16+8y=160+8y+y2AQ = \sqrt {{{( - 10 - 2)}^2} + {{(y - ( - 4))}^2}} = \sqrt {{{(12)}^2} + {{(y + 4)}^2}} = \sqrt {144 + {y^2} + 16 + 8y} = \sqrt {160 + 8y + {y^2}}
Now, we know that point A is equidistant from P and Q so
AP=AQ 145=y2+8y+160AP = AQ \\\ \sqrt {145} = \sqrt {{y^2} + 8y + 160}
Squaring both sides, we get –
145=y2+8y+160 ⇒y2+8y+15=0 ⇒y2+3y+5y+15=0 ⇒y(y+3)+5(y+3)=0 ⇒(y+5)(y+3)=0 ⇒y=−5 or y=−3145 = {y^2} + 8y + 160 \\\ \Rightarrow {y^2} + 8y + 15 = 0 \\\ \Rightarrow {y^2} + 3y + 5y + 15 = 0 \\\ \Rightarrow y(y + 3) + 5(y + 3) = 0 \\\ \Rightarrow (y + 5)(y + 3) = 0 \\\ \Rightarrow y = - 5\,or\,y = - 3
PQ=(−10−3)2+(y−8)2 ⇒PQ=169+(y−8)2 At y=−5, ⇒PQ=169+(−5−8)2=169+169=132units At y=−3, ⇒PQ=169+(−3−8)2=169+121=290  units  PQ = \sqrt {{{( - 10 - 3)}^2} + {{(y - 8)}^2}} \\\ \Rightarrow PQ = \sqrt {169 + {{(y - 8)}^2}} \\\ At\,y = - 5, \\\ \Rightarrow PQ = \sqrt {169 + {{( - 5 - 8)}^2}} = \sqrt {169 + 169} = 13\sqrt 2 units \\\ At\,y = - 3, \\\ \Rightarrow PQ = \sqrt {169 + {{( - 3 - 8)}^2}} = \sqrt {169 + 121} = \sqrt {290} \;units \;
So, the correct answer is “y=-3 OR y=5”.

Note : When two points are equidistant from a single point, it means that the distance between each of the points and the given point is equal that’s why we equated the distance between the points A and P with the distance between the points A and Q. When we know the coordinates of two points in the Cartesian system, we can find out the distance between them by using the distance formula. This formula is derived using the Pythagoras theorem. The Pythagoras theorem simply relates the sides of a right-angled triangle so that if two sides are known, the third side can be calculated easily.