The area of the triangle formed by the graphs of the equations x = 4, y = 3 and 3x + 4y = 12 is

x = 4 ⇒ equation of a line parallel to y - axis. y = 3 ⇒ Equation of a line parallel to x-axis. Putting x = 0 in the equation 3x + 4y = 12, 3 × 0 + 4y = 12 $\Rightarrow y=\frac{12}{4}=3$ ∴ Co-ordinates of the point of intersection on y-axis = (0, 3) Again putting y=0 in the equation 3x+4y= 12, 3x + 4 × 0 = 12 $\Rightarrow x=\frac{12}{3}=4$ ∴ Co-ordinates of the point of intersection on x-axis = (4, 0)

AC = 3 units, BC = 4 units

$\therefore \quad$ Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AC}$

$\quad=\frac{1}{2} \times 4 \times 3=6$ sq. units