find tha abscissa of the point whose ordinate is 4 and is at a distance of 5 units from (7,8).
Answers
Answered by
7
Let the point be
P (x,4)
whose distance from (7,8) is 5 units
√[(7-x)^2 + (8-4)^2] = 5
49+x^2 -14x + 16 = 25
x^2 -14x +40 = 0
x^2 -10x -4x +40 = 0
x (x-10) -4(x-10) = 0
(x-4) (x-10) = 0
x = 4 & 10
------------------------------
Hence the abscissa of the point will be +4 & +10
P (x,4)
whose distance from (7,8) is 5 units
√[(7-x)^2 + (8-4)^2] = 5
49+x^2 -14x + 16 = 25
x^2 -14x +40 = 0
x^2 -10x -4x +40 = 0
x (x-10) -4(x-10) = 0
(x-4) (x-10) = 0
x = 4 & 10
------------------------------
Hence the abscissa of the point will be +4 & +10
Similar questions