1. **Problem:** Find the slope of a line perpendicular to the line whose gradient is $-\frac{3}{4}$. **Formula:** The slope of a line perpendicular to a line with slope $m$ is $-\frac{1}{m}$. **Solution:** Given $m = -\frac{3}{4}$, the perpendicular slope is $-\frac{1}{-\frac{3}{4}} = \frac{4}{3}$. 2. **Problem:** Find the gradient of the line parallel to $3x - 5y + 7 = 0$. **Formula:** For a line $Ax + By + C = 0$, slope $m = -\frac{A}{B}$. Parallel lines have the same slope. **Solution:** $m = -\frac{3}{-5} = \frac{3}{5}$. 3. **Problem:** Write the slope of the line perpendicular to $7x + 6y = 1$. **Solution:** Slope of given line $m = -\frac{7}{6}$. Perpendicular slope $= -\frac{1}{m} = -\frac{1}{-\frac{7}{6}} = \frac{6}{7}$. 4. **Problem:** Find the gradient of the line orthogonal to $5x + 6y = 0$. **Solution:** Slope of given line $m = -\frac{5}{6}$. Orthogonal slope $= -\frac{1}{m} = \frac{6}{5}$. 5. **Problem:** Write the condition for two lines $y = m_1x + c_1$ and $y = m_2x + c_2$ to be parallel and orthogonal. **Solution:** Parallel if $m_1 = m_2$. Orthogonal if $m_1 \times m_2 = -1$. 6. **Problem:** Write the expression to find the angle $\theta$ between two lines $y = m_1x + c_1$ and $y = m_2x + c_2$. **Formula:** $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ 7. **Problem:** Show lines $2x - 8y + 6 = 0$ and $3x - 12y - 4 = 0$ are parallel. **Solution:** Slopes: $m_1 = -\frac{2}{-8} = \frac{1}{4}$, $m_2 = -\frac{3}{-12} = \frac{1}{4}$. Since $m_1 = m_2$, lines are parallel. 8. **Problem:** Find slope of line parallel to $x \cos \alpha + y \sin \alpha = p$. **Solution:** Rewrite as $y = -\frac{\cos \alpha}{\sin \alpha} x + \frac{p}{\sin \alpha}$. Slope $= -\cot \alpha$. 9. **Problem:** Find gradient of line perpendicular to $x \cos \alpha + y \sin \alpha = p$. **Solution:** Perpendicular slope $= \tan \alpha$. 10. **Problem:** Find angle between lines $2x - y + 4 = 0$ and $3x + y + 3 = 0$. **Solution:** Slopes $m_1 = 2$, $m_2 = -3$. $$\tan \theta = \left| \frac{2 - (-3)}{1 + 2 \times (-3)} \right| = \left| \frac{5}{1 - 6} \right| = 1$$ $$\theta = \arctan(1) = 45^\circ$$ Angle between lines is $45^\circ$, so the other angle is $180^\circ - 45^\circ = 135^\circ$. 11. **Problem:** Find acute angle between $3x + y = 10$ and $x + 2y = 6$. **Slopes:** $m_1 = -3$, $m_2 = -\frac{1}{2}$. $$\tan \theta = \left| \frac{-3 - (-\frac{1}{2})}{1 + (-3)(-\frac{1}{2})} \right| = \left| \frac{-\frac{5}{2}}{1 + \frac{3}{2}} \right| = \frac{5/2}{5/2} = 1$$ $$\theta = 45^\circ$$ 12. **Problem:** Show line joining points $(3, -4)$ and $(-2, 6)$ is parallel to $2x + y + 3 = 0$. **Slope of line joining points:** $m = \frac{6 - (-4)}{-2 - 3} = \frac{10}{-5} = -2$. **Slope of given line:** $m = -\frac{2}{1} = -2$. Lines are parallel. 13. **Problem:** Show line joining points $(3, 6)$ and $(-2, -8)$ is perpendicular to $5x + 14y = -3$. **Slope of line joining points:** $m_1 = \frac{-8 - 6}{-2 - 3} = \frac{-14}{-5} = \frac{14}{5}$. **Slope of given line:** $m_2 = -\frac{5}{14}$. Check $m_1 \times m_2 = \frac{14}{5} \times -\frac{5}{14} = -1$. Lines are perpendicular. 14. **Problem:** Find altitude of triangle ABC with vertices $A(3,6)$, $B(-2,1)$, $C(5,-1)$ from vertex A to side BC. **Step 1:** Find slope of BC: $m_{BC} = \frac{-1 - 1}{5 - (-2)} = \frac{-2}{7}$. **Step 2:** Equation of BC: $y - 1 = -\frac{2}{7}(x + 2)$. **Step 3:** Slope of altitude from A is perpendicular to BC: $m_{alt} = \frac{7}{2}$. **Step 4:** Equation of altitude: $y - 6 = \frac{7}{2}(x - 3)$. **Step 5:** Find intersection point D of altitude and BC by solving equations. **Step 6:** Calculate distance AD using distance formula. 15. **Problem:** Prove if $l_1x + m_1y + n_1 = 0$ is parallel to $l_2x + m_2y + n_2 = 0$, then $l_1m_2 - l_2m_1 = 0$. **Proof:** Parallel lines have equal slopes: $-\frac{l_1}{m_1} = -\frac{l_2}{m_2} \Rightarrow l_1m_2 = l_2m_1$. Rearranged: $l_1m_2 - l_2m_1 = 0$. 16. **Problem:** Prove if two lines $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ are perpendicular, then $a_1a_2 + b_1b_2 = 0$. **Proof:** Slopes $m_1 = -\frac{a_1}{b_1}$, $m_2 = -\frac{a_2}{b_2}$. Perpendicular condition: $m_1 m_2 = -1$. Substitute: $\left(-\frac{a_1}{b_1}\right) \left(-\frac{a_2}{b_2}\right) = -1 \Rightarrow \frac{a_1 a_2}{b_1 b_2} = -1$. Multiply both sides by $b_1 b_2$: $a_1 a_2 = -b_1 b_2 \Rightarrow a_1 a_2 + b_1 b_2 = 0$. 17. **Problem:** Find slope of line perpendicular to line joining points $(3, -5)$ and $(-1, 2)$. **Slope of line joining points:** $m = \frac{2 - (-5)}{-1 - 3} = \frac{7}{-4} = -\frac{7}{4}$. **Perpendicular slope:** $-\frac{1}{m} = -\frac{1}{-\frac{7}{4}} = \frac{4}{7}$. 18. **Problem:** Find gradient of line perpendicular to line whose equation is $3x - ...$ (incomplete). **Note:** Problem incomplete, cannot solve. **Final answers summary:** 1) $\frac{4}{3}$ 2) $\frac{3}{5}$ 3) $\frac{6}{7}$ 4) $\frac{6}{5}$ 5) Parallel: $m_1 = m_2$, Orthogonal: $m_1 m_2 = -1$ 6) $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ 7) Parallel (both slopes $\frac{1}{4}$) 8) $-\cot \alpha$ 9) $\tan \alpha$ 10) $135^\circ$ 11) $45^\circ$ 12) Parallel 13) Perpendicular 14) Altitude found by perpendicular distance from A to BC 15) $l_1 m_2 - l_2 m_1 = 0$ 16) $a_1 a_2 + b_1 b_2 = 0$ 17) $\frac{4}{7}$ 18) Incomplete problem, no solution.