Find the values of 'k' if …

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Sia ? 6 years, 4 months ago
Given points will be collinear, if area of the triangle formed by them is zero.
Area = {tex}\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}
{tex}\Rightarrow 0 = \frac { 1 } { 2 } {/tex}[(k + 1)(2k + 3 - 5k) + 3k(5k - 2k) + (5k - 1) (2k - 2k - 3)]
{tex}\Rightarrow {/tex} 0 = (k + 1)(3 - 3k) + 3k(3k) +(5k - 1)(-3)
{tex}\Rightarrow {/tex} 0 = 3k - 3k2 + 3 - 3k + 9k2 - 15k + 3
{tex}\Rightarrow {/tex}0 = 6k2 - 15k + 6
{tex}\Rightarrow {/tex}0 = 2k2 - 5k + 2
{tex}\Rightarrow {/tex}0 = 2k2 - 4k - k + 2
{tex}\Rightarrow {/tex}0 = 2k(k - 2) - 1(k - 2)
{tex}\Rightarrow {/tex}0 = (2k - 1)(k - 2)
{tex}\Rightarrow {/tex}2k - 1 = 0 or k - 2 = 0
{tex}\Rightarrow k = \frac { 1 } { 2 }{/tex} Or k = 2
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