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Question
Find the coordinates of points whose abscissa is −4 and which are at a distance of 15 units from (5, −9).
Sum
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Solution
Coordinates of points whose abscissa = −4 and which are at a distance of 15 units from the point (5,−9).
Step 1: Let the required point be
P(−4, y)
Step 2: Apply the distance formula
`"Distance" = sqrt((x_2 − x_1)^2 + (y_2 − y_1)^2)`
P(−4, y)
Q(5,−9)
Distance = 15
`sqrt((5 − (−4))^2 + (−9 − y)^2) = 15`
`sqrt((9)^2 + (y + 9)^2) = 15`
`sqrt(81 + (y + 9)^2) = 15`
Step 3: Square both sides
81 + (y + 9)2 = 225
(y + 9)2 = 225 − 81
`(y + 9)^2 = 144`
Step 4: Solve for y
y + 9 = ± 12
y + 9 = 12
y = 3
y + 9 = −12
y = −21
The required point are (−4, 3) and (−4, −21)
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