Question

find tha abscissa of the point whose ordinate is 4 and is at a distance of 5 units from (7,8).

Answer 1

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
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Hence the abscissa of the point will be +4 & +10