Question

find the poinys on x axis which are at a distance of 2√5 units from the point (7,-4) how many such points are there?​

Answer 1

Answer: x =  5 and 9

Step-by-step explanation:

Co-ordinates of point on x-axis be (x, 0) which is at a distance of 2√5 from point (7, –4).

Distance between two points A(x1, y1) and B(x2, y2) is AB =  ( x 2 - x 1 ) 2 + ( y 2 - y 1 ) 2  

It is given that the distance between points (x, 0) and (7, –4) is 2√5.

⇒ [ ( 7 -x )2 + ( -4 – 0)2 ] = 2√5 .

Squaring both the sides, we obtain

⇒ ( 7 – x)2 + 16 = 20

⇒  49 + x2 – 14x + 16 = 20

⇒ x2 – 14x + 45 = 0

⇒ x2 – 5x – 9x + 45 = 0

⇒ x( x – 5) – 9( x – 5) = 0

⇒  ( x – 9) ( x – 5) = 0

∴ x = 5 and 9.

Hence, there are two points i.e., (5, 0) and (9, 0).