Find the area of the triangle whose vertices are $(-5,-1), (3,-5), (5,2)$.

If the area of the triangle whose vertices are $(0, 0), (3, 0), (0, 2)$ is $k {\mathrm{unit}}^2$, find the value of $k$.

Find the value of$\text{'}K\text{'}$ for which the points are collinear.

$(7,-2), (5,1), (3,\mathrm{K})$

Find the value of $\mathrm{K}$ for which the points $(8,1), (K,-4), (2,-5)$ are collinear.

Find the value of $\text{'}\mathrm{K}\text{'}$ in improper fraction for which the points are collinear.

$(\mathrm{K},\mathrm{K}), (2,3), (4,-1)$

If we plot these points $A(1,2), B(-3,4), C(7,-1)$ on the coordinate plane and join them, do we get a triangle or a straight line? Explain.