Draw the graphs of \(x - y = -1\) and \(3x + 2y = 12\) on the same graph paper. Find the coordinates of their point of intersection and the area of the triangle formed by these lines and the Y-axis.

Answer: Intersection (2, 3); area = 5 sq. units

\(x - y = -1\) (\(y = x + 1\)): points \((0, 1)\), \((1, 2)\), \((2, 3)\).

\(3x + 2y = 12\): points \((0, 6)\), \((2, 3)\), \((4, 0)\).

[Figure: Both lines drawn; the first cuts the Y-axis at A(0, 1), the second cuts the Y-axis at B(0, 6), and they meet at P(2, 3). Triangle PAB is shaded.]

The base AB lies on the Y-axis: \(AB = 6 - 1 = 5\) units.

The height is the distance of P from the Y-axis, which is its x-coordinate = 2 units.

Area \(= \dfrac{1}{2} \times 5 \times 2 = 5\) sq. units.

Thinking point: for a triangle with the Y-axis, the height is the x-coordinate of the intersection point, not the y-coordinate.