Question

Find the distance of the following points
from x-axix and y-axis
(6,2); (-8,5); (2,-6); (-2, -3)​

Answer 1

Answer:

let the points A(6,2), B(-8,5), C(2,-6) and D(-2,-3 ) be the cordinates of the points A,B,C and D

let the point on x-axis be P(x,0)

by using distance formula

$\sqrt({X2}-X1)^{2} +(Y2-Y1)^{2}$

AP  = √(x-6)²+(0-2)²

      = √x²+36+12x +4

      = √x²+12x+40

BP  = √(x+8)²+(0-5)²

      = √x²+64+16x+25  

      =  √x²+89+16x

CP  = √(x-2)²+(0+6)²

      = √ x²+4-4x+36

      = √x²+40-4x

DP  = √(x+2)²+(0-3)²

      = √(x²+4+4x)+9

      =  √(x²+13+4x

let the point on y-axis be Q(0,y)

AQ = √(0-6)²+(y-2)²

     = √36+y²+4-4y

     = √y²-4y+40

BQ = √(0+8)²+(y-5)²

     = √64+y²+25-10y

     = √y²+89-10y

CQ = √(0-2)²+(y+6)²

     = √4+y²+36+12y

     = √y²+10+12y

DQ = √(0+2)²+(y+3)²

      = √4+y²+9+6y

      =√y²+6y+13

Hence , above are the distances between the points from x-axis and y-axis