The vertices of the given ΔABC are A(1 ,-1), B( -4, 2k ) and C (-k, 5) 

Area of ΔABC =

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{points} \mathrm{are} \mathrm{A}( -2, 3 ) \mathrm{B}( 8, 3 ) \mathrm{and} \mathrm{C}( 6, 7 ). \\ \mathrm{using} \mathrm{distance} \mathrm{firmula}, \mathrm{we} \mathrm{have}: \\ {\mathrm{AB}}^2 =( 8 - ( -2 ){)}^2 +( 3 - 3 {)}^2 \\ \Rightarrow {\mathrm{AB}}^2 = ( 8+ 2 {)}^2 + 0 \\ \Rightarrow {\mathrm{AB}}^2 = {10}^2 \\ \Rightarrow {\mathrm{AB}}^2 = 100 \\ {\mathrm{BC}}^2 = (6-8 {)}^2 + (7-3{)}^2 \\ \Rightarrow {\mathrm{BC}}^2 = ( - 2 {)}^2 + 4^2 \\ \Rightarrow {\mathrm{BC}}^2 = 4+16 \\ \Rightarrow {\mathrm{BC}}^2 = 20 \\ {\mathrm{CA}}^2 = ( -2 -6 {)}^2 + ( 3-7 {)}^2 \\ \Rightarrow {\mathrm{CA}}^2 = (-8{)}^2 +(-4 {)}^2 \\ \Rightarrow {\mathrm{CA}}^2 = 64 + 16 \\ \Rightarrow {\mathrm{CA}}^2 =80 \\ \mathrm{It} \mathrm{can} \mathrm{be} \mathrm{observe} \mathrm{that}: \\ {\mathrm{BC}}^2 + {\mathrm{CA}}^2 = 20 + 80 =100 = {\mathrm{AB}}^2 \\ \mathrm{So}, \mathrm{by} \mathrm{the} \mathrm{converse} \mathrm{of} \mathrm{pythagoras} \mathrm{Theorem}, \\ ∆\mathrm{ABC} \mathrm{is} \mathrm{a} \mathrm{right} \mathrm{angled} \mathrm{triangle} \mathrm{with} \mathrm{right} \mathrm{angled} \mathrm{at} \mathrm{C}. \end{gathered}$$

Let A(a, a2), B(b, b2) and C(0, 0) be the coordinates of the given points.
We know that the area of a triangle having vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is ∣∣12[x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)]∣∣ square units.
So,
Area of ∆ABC\

Since the area of the triangle formed by the points (a, a²), (b, b²) and (0, 0) is not zero, so the given points are not collinear.

$$\begin{gathered} \mathrm{The} \mathrm{given} \mathrm{vertises} \mathrm{are} \mathrm{A}(\mathrm{x},\mathrm{y}), \mathrm{B}(1,2) \mathrm{C}(2,1). \\ \mathrm{It} \mathrm{is} \mathrm{know} \mathrm{that} \mathrm{the} \mathrm{area} \mathrm{of} \mathrm{a} \mathrm{tringle} \mathrm{whose} \mathrm{vertises} \mathrm{are} ({\mathrm{x}}_1,{\mathrm{y}}_1). ({\mathrm{x}}_{2,} {\mathrm{y}}_2 ), \\ \mathrm{and} ({\mathrm{x}}_{3, }{\mathrm{y}}_{3) }\mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{1}{2} \left|{\mathrm{x}}_1 ({\mathrm{y}}_2-{\mathrm{y}}_{3 }) + {\mathrm{x}}_{2 }({\mathrm{y}}_3-{\mathrm{y}}_1) + {\mathrm{x}}_3({\mathrm{y}}_1-{\mathrm{y}}_2)\right| \\ \therefore \mathrm{Area} \mathrm{of} ∆\mathrm{ABC} \\ = \frac{1}{2} \left|\mathrm{x}(2-1) + 1(1-\mathrm{y}) +2(\mathrm{y}-2)\right| \\ =\frac{1}{2} \left|\mathrm{x}+1-\mathrm{y}+2\mathrm{y}-4\right| \\ =\frac{1}{2} \left|\mathrm{x}+\mathrm{y}-3\right| \\ \mathrm{The} \mathrm{area} \mathrm{of} ∆\mathrm{ABC} \mathrm{is} \mathrm{given} \mathrm{as} 6 \mathrm{sq}.\mathrm{units}. \\ \Rightarrow \frac{1}{2}\left|\mathrm{x}+\mathrm{y}-3\right| = 6 \\ \Rightarrow \mathrm{x}+\mathrm{y}-3 = 12 \\ \therefore \mathrm{x}+\mathrm{y}=15 \end{gathered}$$