Question

find the points on the x axis which are at a distance of 5 units from (3,4)​

Answer 1

Step-by-step explanation:

Given the point is on x axis. So y−coordinate is 0.

Let the points on X−axis be A(x,0) which is at a distance of 5 units from B(5,−4).

By distance formula , distance between AB=

[(x

2

−x

1

)

2

+(y

2

−y

1

)

2

]

5=

[(5−x)

2

+(−4−0)

2

]

5=

[(5−x)

2

+−4

2

]

5=

[(25+x

2

−10x+16)]

5=

[(x

2

−10x+41)]

Squaring both sides

25=x

2

−10x+41

x

2

−10x+41−25=0

x

2

−10x+16=0

x

2

−2x−8x+16=0

(x−2)(x−8)=0

x−2=0 or x−8=0

x=2 or x=8

Hence the points are (2,0) and (8,0