#### notes

Here what will be the distance between points PQ? To find this we will use distance formula.

Distance formula says, PQ= `sqrt[ (x_2-x_1)^2 + (y_2-y_1)^2]`

Example1- Find the distance between the points P(6,-6) and O(0,0)

Solution: `x_2=0, x_1=6, y_2=0 and y_1=-6`

PO= `sqrt[ (x_2-x_1)^2 + (y_2-y_1)^2]`

= `sqrt[ (0-6)^2 + (0+6)^2]`

= `sqrt[ (-6)^2 + (6)^2]`

= `sqrt[ 36+36]`

= `sqrt72`

PO = 6 `sqrt2` units

Example2- Find the values of x for which the distance between the points P(4,-5) and Q(12,x) is 10 units

Solution: PQ=10

`x_2= 12, x_1=4, y_2=x and y_1=-5`

PQ= `sqrt[ (x_2-x_1)^2 + (y_2-y_1)^2]`

10 = `sqrt[ (12-4)^2 + (x+5)^2]`

= `sqrt[ (8)^2 + (x+5)^2]`

= `sqrt[ 64 + (x+5)^2]`

Sqauring both the sides

`100= 64+ (x+5)^2`

`36= (x+5)^2`

`+or-6 = x+5`

`x= 6-5 or x= -6-5`

`x=1 or x= -11`

Example3- Find the point P on x-axis which is equidistance from the points A(5,-2) and B(-3,2)

Solution: Let the point be P(x,0)

Now, PA=PB [Given as point P on x-axis is equidistance from the points A(5,-2) and B(-3,2)]

`"PA"^2="PB"^2`

`(x_2-x_1)^2 + (y_2-y_1)^2 = (x_2-x_1)^2 + (y_2-y_1)^2`

`(5-x)^2 + (-2-0)^2 = (-3-x)^2 + (2-0)^2`

`25+x^2-10x+4 = (-3)^2 +x^2 -2(-3)(x) +4`

`25+x^2-10x+4 = 9+x^2+6x+4`

`16x= 16`

`x=1`

Therefore, point P is (1,0)

#### Shaalaa.com | Coordinate Geometry part 2 (Distance Formula)

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