Find the value of 'k' vertices …
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Sia ? 4 years, 9 months ago
Let the points A (2, 3), B{tex}\left( {4,k} \right){/tex} and C{tex}\left( {6, - 3} \right){/tex} be collinear.
If the points are collinear then area of triangle ABC formed by these three points is 0.
{tex}\therefore {/tex}{tex}{\text{ar}}\left( {\Delta {\text{ABC}}} \right) = \frac{1}{2}\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right]{/tex}= 0
{tex} \Rightarrow {/tex}{tex}\frac{1}{2}\left[ {2\left( {k + 3} \right) + 4\left( { - 3 - 3} \right) + 6\left( {3 - k} \right)} \right] = 0{/tex}
{tex} \Rightarrow {/tex}{tex}\left[ {2k + 6 - 24 + 18 - 6k} \right] = 0{/tex}
{tex} \Rightarrow {/tex}{tex}\left[ { - 4k} \right] = 0{/tex}
{tex} \Rightarrow {/tex}{tex}k = 0{/tex}
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