Question

It the points [7.-2] [5.1] . [3.k]Jare
Collinear find the volue
of 'k!​

Answer 1

$\bf \underline{ \underline{\maltese\:Given} }$

$\sf The \: points \: (7,-2) , (5,1) and (3,k) \: are \: collinear.$

$\bf \underline{\underline{\maltese\: To \: find }}$

$\sf \implies Value \: of \: K = \: ?$

$\bf \underline{\underline{\maltese\: Solution }}$

$\sf If \: the \: points \: are \: Collinear \: then,$

$\sf \implies Area \: of \: triangle = 0$

$\sf Let \: the \: points \: be :$

$\bf \implies A(7,-2) , \: B(5,1) \: and \: C(3,k)$

$\bf \underline{Now},$

$\underline{\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times | \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ |}}$

$\bf \underline{Where},$

$\bullet \implies \: \sf x_1 = 7$

$\bullet\implies \: \sf x_2 = 5$

$\bullet\implies \: \sf x_3 = 3$

$\bullet\implies \: \sf y_1 = - 2$

$\bullet \implies \: \sf y_2 = 1$

$\bullet \implies \: \sf y_3 = K$

$\sf Now, substituting \: the \: values,$

$\sf \implies \dfrac{1}{2} \times | \ 7(1-K )+5(K - (- 2))+3( - 2-1) \ | = 0$

$\sf \implies \dfrac{1}{2} \times | \ 7(1-K )+5(K + 2)+3( - 2-1) \ | = 0$

$\sf \implies \dfrac{1}{2} \times | \ 7(1-K )+5(K + 2)+3 (- 3)\ | = 0$

$\sf \implies\dfrac{1}{2} \times | \ 7 - 7k + 5k + 10 - 9 \ | = 0$

$\sf \implies \dfrac{1}{2} \times | \ 8 - 2k \ |= 0$

$\sf \implies 8 - 2k = \cancel{0 \times \bigg(\dfrac{1}{2} \bigg) }$

$\sf \implies 8 - 2k = 0$

$\sf \implies 8 = 2k$

$\sf \implies 4 = k$

$\underline{ \boxed{ \red{ \bf Therefore, the \: value \: of \: K = 4 }}}$