Question

4. Find the co-ordinates of points on the x-axis
which are at a distance of 17 units from the
point (11, -8)​

Answer 1

Answer:

Step-by-step explanation:

ANY POINT ON THE X AXIS IS OF THE FORM (x,0)

x1 = 11

x2 = x

y1 = -8

y2 = 0

squaring on both sides, the distance formula becomes

(x2 - x1)^2  + (y2 - y1)^2 = 17^2

(x - 11)^2  +    (0-{-8})^2 = 289

x^2 - 22x + 121 + 64 = 289

x^2 - 22x = 289 - 185

x^2 - 22x = 104

x^2 - 22x - 104 = 0

x^2 - 26x + 4x -104      (by factorizing)

x (x - 26)  + 4 (x - 26)

(x -26)  (x+4)

x=26  (or)   x=-4

therefore, the required points are (26,0)  and  (-4,0)