Question

find the value of kfor which the points are collinear.(7,-2)( 5,1)(3,k)​

Answer 1

$\large \green{ \underline{answer}} \\ \\ \large \implies{k \: = \: 4} \\ \\ \large \implies \orange{ \underline{ \underline{step \: . \: by \: . \: step \: . \: explanation}}} \\ \\ \large \implies{let \: X1 = 7 \: \: X2 = 5 \: \: X3 = 3} \\ \\ \large \implies{let \: Y1 = - 2 \: \: Y2 = 1 \: \: Y3 = k} \\ \\ \large \implies{if \: \: point \: \: are \: \: collinear \:} \\ \\ \large \implies{ area \: \: of \: \triangle \: abc \: = 0} \\ \\ \large \implies{ \frac{1}{2} (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) = 0} \\ \\ \large \implies{ \frac{1}{2} (7(1 - k) + 5(k + 2) + 3( - 2 - 1) = 0} \\ \\ \large \implies{7 - 7k + 5k + 10 - 9 = 0} \\ \\ \large \implies{ - 7k + 5k = - 10 + 9 - 7} \\ \\ \large \implies{ - 2k = - 8} \\ \\ \large \implies \orange{ \boxed{k = 4}}$