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Question
In fig., PS = 2, SQ = 6, QR = 5, PT = x and TR = y. Then find the pair of value of x and y such that ST || side QR.
Sum
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Solution
In ΔPQR,
Line ST || side QR ...[Given]
∴ `"PS"/"SQ" = "PT"/"TR"` ...[Basic proportionality theorem]
∴ `2/6 = "PT"/"TR"` ...[Given]
∴ `1/3 = x/y`
∴ y = 3x
When x = 1, y = 3(1) = 3
∴ x = 1, y = 3
When x = 2, y = 3(2) = 6
∴ x = 2, y = 6
When x = 3, y = 3(3) = 9
∴ x = 3, y = 9
∴ Some of the pairs of values of x and y are (1, 3), (2, 6), (3, 9).
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