7,-4). How many16. Find the points on the x-axis which are at a distan

1.	7,-4). How many16. Find the points on the x-axis which are at a distance of 2 v5 from the point (7, -4). How manysuch points are there ?
Answer» Co-ordinates of point onx-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 =(x2-x1)^2+(y2-y1)^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 + x^2– 14x + 16 = 20 ⇒ x^2– 14x + 45 = 0 ⇒ x^2– 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).

7,-4). How many16. Find the points on the x-axis which are at a distance of 2 v5 from the point (7, -4). How manysuch points are there ?

Answer»

Co-ordinates of point onx-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 =(x2-x1)^2+(y2-y1)^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 + x^2– 14x + 16 = 20

⇒ x^2– 14x + 45 = 0

⇒ x^2– 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).