For what value of x are …

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Sia ? 6 years, 6 months ago
According to the question, A(-3, 12), B(7, 6) and C(x, 9).
(x1 = -3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
The given points A,B,C are collinear. So, area of triangle is zero.
{tex}\Rightarrow{/tex} x1(y2 - y3) + x2(y3 - y1)+ x3(y1 - y2)=0
{tex}\Rightarrow{/tex} {tex}(-3)(6 - 9) + 7 (9 - 12) +x(12 - 6) =0{/tex}
{tex}\Rightarrow{/tex} {tex}9 - 21 + 6x = 0{/tex}
{tex}\Rightarrow{/tex} {tex}6x = 12 {/tex}
{tex}\Rightarrow{/tex} {tex}x = 2{/tex}
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