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Question
Show that the point (0, 9) is equidistant from the points (– 4, 1) and (4, 1)
Solution
Let P(x1, y1) = P(0, 9), Q(x2, y2) = Q(– 4, 1), R(x3, y3) = R(4, 1)
By distance formula,
d(P, Q) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`
= `sqrt([(-4) - 0]^2 + (1 - 9)^2`
= `sqrt((-4)^2 + (-8)^2`
= `sqrt(16 + 64)`
= `sqrt(80)`
= `4sqrt(5)`
And
d(P, R) = `sqrt((x_3 - x_1)^2 + (y_3 - y_1)^2`
= `sqrt((4 - 0)^2 + (1 - 9)^2`
= `sqrt(4^2 + (-8)^2`
= `sqrt(16 + 64)`
= `sqrt(80)`
= `4sqrt(5)`
Here, d(P, Q) = d(P, R)
∴ The point (0, 9) is equidistant from (– 4, 1) and (4, 1).
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By distance formula,
PQ = `sqrt(square + (y_2 - y_1)^2`
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= `5sqrt(5)`
