If the point \[A(2, - 4)\] is equidistant from $ P(3,8) $ and $ Q( - 10,y) $ . Find the value of y. Also find PQ.
Answer
637.2k+ views
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 = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Distance between \[A(2, - 4)\] and $ P(3,8) $ is
$ AP = \sqrt {{{(3 - 2)}^2} + {{(8 - ( - 4))}^2}} = \sqrt {{{(1)}^2} + {{(12)}^2}} = \sqrt {145} $
Distance between \[A(2, - 4)\] and $ Q( - 10,y) $ is
$ AQ = \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 \\
\sqrt {145} = \sqrt {{y^2} + 8y + 160} $
Squaring both sides, we get –
$ 145 = {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 = \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.
Complete step-by-step answer:
Distance of between two points is given by the formula,
$ distance = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Distance between \[A(2, - 4)\] and $ P(3,8) $ is
$ AP = \sqrt {{{(3 - 2)}^2} + {{(8 - ( - 4))}^2}} = \sqrt {{{(1)}^2} + {{(12)}^2}} = \sqrt {145} $
Distance between \[A(2, - 4)\] and $ Q( - 10,y) $ is
$ AQ = \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 \\
\sqrt {145} = \sqrt {{y^2} + 8y + 160} $
Squaring both sides, we get –
$ 145 = {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 = \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.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

