Line Gradient 5B93A0

1. **State the problem:** We are given two points $A(2, -3)$ and $B(7, k)$, and the gradient of the line passing through them is $\frac{3}{2}$. We need to find $k$ and then show that the perpendicular bisector of $AB$ has the equation $8x + 12y - 45 = 0$. 2. **Find the value of $k$ using the gradient formula:** The gradient $m$ of a line through points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Here, $m = \frac{3}{2}$, $x_1 = 2$, $y_1 = -3$, $x_2 = 7$, and $y_2 = k$. 3. **Set up the equation:** $$\frac{k - (-3)}{7 - 2} = \frac{3}{2}$$ Simplify the denominator: $$\frac{k + 3}{5} = \frac{3}{2}$$ 4. **Cross multiply to solve for $k$:** $$2(k + 3) = 3 \times 5$$ $$2k + 6 = 15$$ 5. **Isolate $k$:** $$2k = 15 - 6$$ $$2k = 9$$ $$k = \frac{9}{2}$$ 6. **Value of $k$ is:** $$k = 4.5$$ --- 7. **Find the midpoint $M$ of $AB$:** Midpoint formula: $$M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$$ Substitute values: $$M = \left( \frac{2 + 7}{2}, \frac{-3 + 4.5}{2} \right) = \left( \frac{9}{2}, \frac{1.5}{2} \right) = \left(4.5, 0.75\right)$$ 8. **Find the gradient of $AB$ again:** Already given as $\frac{3}{2}$. 9. **Gradient of perpendicular bisector:** The perpendicular bisector is perpendicular to $AB$, so its gradient $m_{\perp}$ satisfies: $$m \times m_{\perp} = -1$$ $$\frac{3}{2} \times m_{\perp} = -1$$ $$m_{\perp} = -\frac{2}{3}$$ 10. **Equation of perpendicular bisector:** Using point-slope form: $$y - y_1 = m_{\perp}(x - x_1)$$ Substitute $M(4.5, 0.75)$ and $m_{\perp} = -\frac{2}{3}$: $$y - 0.75 = -\frac{2}{3}(x - 4.5)$$ 11. **Expand and simplify:** $$y - 0.75 = -\frac{2}{3}x + 3$$ $$y = -\frac{2}{3}x + 3 + 0.75$$ $$y = -\frac{2}{3}x + 3.75$$ 12. **Rewrite in standard form:** Multiply both sides by 3 to clear denominator: $$3y = -2x + 11.25$$ Bring all terms to one side: $$2x + 3y - 11.25 = 0$$ 13. **Multiply entire equation by 4 to clear decimal:** $$4(2x + 3y - 11.25) = 0$$ $$8x + 12y - 45 = 0$$ 14. **Conclusion:** The perpendicular bisector of $AB$ has the equation: $$8x + 12y - 45 = 0$$ which matches the required equation. **Final answers:** - $k = 4.5$ - Perpendicular bisector equation: $8x + 12y - 45 = 0$