The area of the triangle formed by the points (k, 2 – 2k), (– k + 1, 2k) and (– 4 – k, 6 – 2 k) is 70 units. For

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four real values of k

no integral value of k

two integral values of k

only one integral value of k

answer is D.

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Detailed Solution

k2−2k1−k+12k1−4−k6−2k1=140which is true for only one integral value of k.

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