Question

find the value of k for which the points with coordinates bracket 2, 5 bracket close, bracket 4, 6 bracket close bracket ke, 11 upon 2 bracket close are collinear

Answer 1

$hello$

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$\frac{1}{2} \times {7(1 - k) + 5k - ( - 2)) + 3( - 2 - 1)} = 0$

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$7 - 7k + 5k + 10 - 9 = 0$

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$- 2k = - 8$

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$k = \frac{ - 8}{ - 2}$

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$k = 4$

$<b>$

†thankS†

Answer 2

$<b>$

★hello mate★

↩here us ur anSwer↪

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$\frac{1}{2} \times {7(1 - k) + 5k - ( - 2)) + 3( - 2 - 1)} = 0$

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$7(1 - k) + 5(k + 2) + 3( -3 ) = 0 \times 2$

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$7 - 7k + 5k + 10 - 9 = 0$

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$- 2k = - 8$

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$k = \frac{ - 8}{ - 2}$

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$k = 4$

$<b>$

†thankS†