Question

find the point on x axis which are at a distance of 2 root 5 units from the point [7,-4] . how many such points are there

Answer 1

let coordinates of the point=(x,0)(as the poin lies on x axis)
x1=7. y1=-4
x2=x. y2=0
distance between 2 pts=√(x2-x1)^2+(y2-y1)^2
A/Q.....
2√5=√(x-7)^2+(0-(-4))^2
on squaring both sides....
20=x^2+49-14x+16
20=x^2+65-14x
0=x^2-14x+45
0=x^2-9x-5x+45
0=x(x-9)-5(x-9)
0=(x-9)(x-5)
x-9=0. x-5=0
x=9. x=5
therefore x=9,5
and coordinates of points.....(9,0)or(5,0)

Answer 2

Answer:x=9,5

Step-by-step explanation:

let coordinates of the point=(x,0)(as the point lies on x axis)

x1=7. y1=-4

x2=x. y2=0

distance between 2 pts=√(x2-x1)^2+(y2-y1)^2

A/Q.....

2√5=√(x-7)^2+(0-(-4))^2

on squaring both sides....

20=x^2+49-14x+16

20=x^2+65-14x

0=x^2-14x+45

0=x^2-9x-5x+45

0=x(x-9)-5(x-9)

0=(x-9)(x-5)

x-9=0. x-5=0

x=9. x=5

therefore x=9,5

and coordinates of points.....(9,0)or(5,0)