Question

a) The distance between the points P(1, 3) and Q(x, 7) is 5 units. Find the possible values of x.

Answer 1

Step-by-step explanation:

p(1,3) Q(x,7)

distance=√(x2-x1)²+(y2-y1)²

5=√(x-1)²+(7-3)²

5=√x²+1-2x+16

25=x²-2x+17

x²-2x-8=0

x²-4x+2x-8=0

x(x-4)+2(x-4)=0

(x+2)(x-4)=0

x=-2,4

Answer 2

Answer:

x=(4,-2)

Step-by-step explanation:

using formula

25=(x-1)^2+(7-3)^2

25-16=x^2-2x+1

9-1=x^2-2x

x^2-2x-8=0

x^2-(4-2)x-8=0

x^2-4x+2x-8=0

x(x-4)+2(x-4)=0

(x-4) (x+2)=0

either                          or

x-4=0                      x+2=0

x=4                           x=-2