Q.Find the value of k if the area of triangle is $\dfrac{33}{2}$ square units and the vertices are L(3,-5), M(-2,k), N(1,4)

Given a triangle's vertices $A(x_1,y_1),B(x_2,y_2),C(x_3,y_3)$, its area is

derived by splitting the triangle's area into three trapeziums against the x-axis and simplifying. Because listing the vertices in a different rotational order swaps two rows of the determinant (Property 2), the signed determinant can come out negative — so the area itself is always taken as the absolute value. This immediately gives a clean collinearity test: if the same determinant evaluates to exactly $0$, the "triangle" has zero area, which can only happen when the three points lie on one straight line. This formula and its collinearity …