1. **Stating the problem:** We have two functions $f(x) = 2x^2 - x - 1$ and $g(x) = x^2 - 3x + 7$ intersecting at two points $K(a,b)$ and $L(c,d)$. A line $m$ passes through these points. 2. **Find intersection points:** Set $f(x) = g(x)$ to find $x$-coordinates of intersection points: $$2x^2 - x - 1 = x^2 - 3x + 7$$ 3. **Simplify the equation:** $$2x^2 - x - 1 - x^2 + 3x - 7 = 0$$ $$x^2 + 2x - 8 = 0$$ 4. **Factor the quadratic:** $$x^2 + 2x - 8 = (x + 4)(x - 2) = 0$$ So, $x = -4$ or $x = 2$. 5. **Find corresponding $y$ values for each $x$ using $f(x)$:** For $x = -4$: $$b = f(-4) = 2(-4)^2 - (-4) - 1 = 2(16) + 4 - 1 = 32 + 4 - 1 = 35$$ For $x = 2$: $$d = f(2) = 2(2)^2 - 2 - 1 = 2(4) - 2 - 1 = 8 - 2 - 1 = 5$$ 6. **Compare $b$ and $d$:** Since $b = 35 > d = 5$, the value of $a$ (the $x$-coordinate of $b$) is $-4$. 7. **Find gradient (slope) of line $m$ passing through $K(a,b)$ and $L(c,d)$:** $$m = \frac{d - b}{c - a} = \frac{5 - 35}{2 - (-4)} = \frac{-30}{6} = -5$$ 8. **Find slope $p$ of line perpendicular to $m$:** The slope of a line perpendicular to slope $m$ is: $$p = -\frac{1}{m} = -\frac{1}{-5} = \frac{1}{5}$$ 9. **Find $q$ using point $(1,1)$ on line $y = px + q$:** $$1 = \frac{1}{5} \times 1 + q \Rightarrow q = 1 - \frac{1}{5} = \frac{4}{5}$$ 10. **Calculate $p + q$:** $$p + q = \frac{1}{5} + \frac{4}{5} = 1$$ **Final answers:** - $a = -4$ - Gradient of line $m = -5$ - $p + q = 1$