Question

find the value of x for which the distance between the points (x,-5) and (2,7) is 13 units.

Answer 1

Distance between the points = √[(x2-x1)^2+(y2-y1)^2]

Say the points (x1,y1) = (x,-5)

(x2,y2) = (2,7)

Distance between the points = √[(2-x)^2+(7+5)^2]

13 = √(4+x^2-4x+144)

13 = √(x^2-4x+148)

Squring on each side.

169 = x^2-4x+148

Bring all terms one side.

x^2-4x+148-169 = 0

x^2-4x-21 = 0

x^2-7x+3x-21 = 0

x(x-7)+3(x-7) = 0

(x-7)+(x+3) = 0

x-7 = 0 and x+3 = 0

x = 7, x= -3