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

Question

5657 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-29T15:08:32

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

The formula for the distance between two points, #(x_2,y_2)# and #(x_1,y_1)# is:

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

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

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

If the point is on the x-axis, then #y_2# must equal 0:

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

Solve for x_2:

#25 = (x_2-2)^2+(0--3)^2#

#25 = (x_2-2)^2+9#

#(x_2-2)^2=16#

#x_2-2=-4# and #x_2-2=4#

#x_2=-2# and #x_2=6#