Asked by Anne-Marie
Find all points on the x-axis that are 10 units from the point (4,-5).
Answers
Answered by
Reiny
let the point be (x,0)
distance = √(x-4)^2 + (0+5)^2 ) = 10
square both sides:
(x-4)^2 + 25 = 100
x^2 - 8x = 75
completing the square
x^2 - 8x + 16 = 75+16
(x-4)^2 = 91
x - 4 = ±√91
x = 4 ± √91
or
consider a circle with centre (4,-5) and radius 10, then find the x-intercepts
(x-4)^2 + (y+5)^2 = 100
for x-intercepts, let y = 0
(x-4)^2 + 25 = 100 , same equation as above
distance = √(x-4)^2 + (0+5)^2 ) = 10
square both sides:
(x-4)^2 + 25 = 100
x^2 - 8x = 75
completing the square
x^2 - 8x + 16 = 75+16
(x-4)^2 = 91
x - 4 = ±√91
x = 4 ± √91
or
consider a circle with centre (4,-5) and radius 10, then find the x-intercepts
(x-4)^2 + (y+5)^2 = 100
for x-intercepts, let y = 0
(x-4)^2 + 25 = 100 , same equation as above
Answered by
[email protected]
X^2=72
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