Advertisement Remove all ads

Distance Formula

Advertisement Remove all ads

Topics

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)

If you would like to contribute notes or other learning material, please submit them using the button below.

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

Shaalaa.com


Next video


Shaalaa.com


Coordinate Geometry part 2 (Distance Formula) [00:10:15]
S
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×