Question

A(3, 4) and B (7, 8) are two points. P (a, b) is equidistant from A and B such that AP = √26. (a) Show that a + b = 11. (b) Find the values of a and b.​

Answer 1

The value of a is 2 and that of b is 9.

What is the distance formula?

The distance between two points is the length of the line joining the two points. If the two points lie on the same horizontal or same vertical line, the distance can be found by subtracting the coordinates that are not the same.

The distance between two points of the xy-plane can be found using the distance formula.

An ordered pair (x, y) represents co-ordinate of the point, where x-coordinate (or abscissa) is the distance of the point from the x-axis and y-coordinate (or ordinate) is the distance of the point from the y-axis.

The distance formula is given as:

d = √($x_{2}$-$x_{1}$)²+($y_{2}$-$y_{1}$)²

In the given question it is said that P(a,b) is equidistant from A and B.

AP is given √26

So we can apply distance formula to show that a+b=11

√(a-3)²+(b-4)²=√(7-a)²+(8-b)²

(a-3)²+(b-4)²=(7-a)²+(8-b)²

a²-6a+9+b²+16+8b=a²-14a+49+b²-16b+64

25-6a-8b=113-14a-16a

88=8a+8b

(a+b)8=88

a+b=11

Hence proved

Now we need to find the value of a and b,

AP=√(a-3)²+(b-4)²

26=(a-3)²+(b-4)²

26=(11-b-3)²+(b-4)².                   As a+b=11 so a=11-b

26=(8-b)²+(b-4)²

26=64-16b+b²+b²-8b+16

26=2b²-24b+80

b²-12b+27=0

b²-9b-3b+27=0

b(b-9)-3(b-9)=0

So the value of b=9

As a=11-b=11-9

So a=2

To learn more about distance formula,

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