Find the coordinate of the point on x-axis which is at a distance of 5 units from the point (2,-3)(2,3)?

1 Answer
Mar 29, 2018

There are two such points, (-2,0)(2,0) and (6,0)(6,0)

Explanation:

The formula for the distance between two points, (x_2,y_2)(x2,y2) and (x_1,y_1)(x1,y1) is:

d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)" [1]"d=(x2x1)2+(y2y1)2 [1]

Substitute (x_1,y_1) = (2,-3)(x1,y1)=(2,3) and d = 5d=5 into equation [1]:

5 = sqrt((x_2-2)^2+(y_2--3)^2)" [1.1]"5=(x22)2+(y23)2 [1.1]

If the point is on the x-axis, then y_2y2 must equal 0:

5 = sqrt((x_2-2)^2+(0--3)^2)" [1.2]"5=(x22)2+(03)2 [1.2]

Solve for x_2:

25 = (x_2-2)^2+(0--3)^225=(x22)2+(03)2

25 = (x_2-2)^2+925=(x22)2+9

(x_2-2)^2=16(x22)2=16

x_2-2=-4x22=4 and x_2-2=4x22=4

x_2=-2x2=2 and x_2=6x2=6