1. **State the problem:** Find the coordinates of the points A and B where the line $y=4x-3$ meets the curve $y=3+5x-2x^2$. Then find the area of triangle $POQ$ formed by the perpendicular bisector of $AB$ intersecting the coordinate axes. 2. **Find points A and B:** Set the line equal to the curve: $$4x - 3 = 3 + 5x - 2x^2$$ Bring all terms to one side: $$0 = 3 + 5x - 2x^2 - (4x - 3)$$ Simplify: $$0 = 3 + 5x - 2x^2 - 4x + 3 = 6 + x - 2x^2$$ Rearranged: $$-2x^2 + x + 6 = 0$$ Multiply by $-1$ to simplify: $$2x^2 - x - 6 = 0$$ 3. **Solve the quadratic:** Using the quadratic formula $x=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ with $a=2$, $b=-1$, $c=-6$: $$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times (-6)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 48}}{4} = \frac{1 \pm \sqrt{49}}{4} = \frac{1 \pm 7}{4}$$ So: $$x_1 = \frac{1 + 7}{4} = 2$$ $$x_2 = \frac{1 - 7}{4} = -\frac{3}{2}$$ 4. **Find y-coordinates:** Using the line $y=4x-3$: For $x=2$: $$y = 4(2) -3 = 8 -3 = 5$$ For $x=-\frac{3}{2}$: $$y = 4 \times (-\frac{3}{2}) - 3 = -6 - 3 = -9$$ Thus, $$A = (2, 5), \quad B = \Big(-\frac{3}{2}, -9\Big)$$ 5. **Find midpoint M of AB:** $$M_x = \frac{2 + (-\frac{3}{2})}{2} = \frac{2 - 1.5}{2} = \frac{0.5}{2} = 0.25$$ $$M_y = \frac{5 + (-9)}{2} = \frac{-4}{2} = -2$$ Midpoint $M = (0.25, -2)$. 6. **Find slope of AB:** $$m_{AB} = \frac{5 - (-9)}{2 - (-\frac{3}{2})} = \frac{14}{2 + 1.5} = \frac{14}{3.5} = 4$$ 7. **Slope of perpendicular bisector:** $$m_{perp} = -\frac{1}{m_{AB}} = -\frac{1}{4}$$ 8. **Equation of perpendicular bisector:** Using point-slope form at $M(0.25, -2)$: $$y - (-2) = -\frac{1}{4}(x - 0.25)$$ Simplify: $$y + 2 = -\frac{1}{4}x + \frac{1}{16}$$ $$y = -\frac{1}{4}x + \frac{1}{16} - 2$$ $$y = -\frac{1}{4}x - \frac{31}{16}$$ 9. **Find intercepts P and Q:** - $P$ on $x$-axis: set $y=0$ $$0 = -\frac{1}{4}x - \frac{31}{16}$$ Multiply both sides by 16: $$0 = -4x - 31$$ $$4x = -31$$ $$x = -\frac{31}{4} = -7.75$$ So $P = (-7.75, 0)$. - $Q$ on $y$-axis: set $x=0$ $$y = -\frac{1}{4} \times 0 - \frac{31}{16} = -\frac{31}{16} = -1.9375$$ So $Q = (0, -1.9375)$. 10. **Calculate area of triangle POQ:** With vertices $P(-7.75, 0)$, $O(0,0)$, $Q(0, -1.9375)$, the triangle is right-angled at $O$, area is: $$\text{Area} = \frac{1}{2} \times |x_P| \times |y_Q| = \frac{1}{2} \times 7.75 \times 1.9375 = \frac{1}{2} \times 15.00390625 = 7.501953125$$ Rounded: $$\approx 7.50$$ **Final answers:** (a) $A = (2, 5)$, $B = (-\frac{3}{2}, -9)$ (b) Area of triangle $POQ$ is approximately $7.50$ units squared.