1. **Find the equation of the line that passes through (4,3) with gradient $g=2$.** The formula for the equation of a line with gradient $m$ passing through point $(x_1,y_1)$ is: $$y - y_1 = m(x - x_1)$$ Substitute $m=2$, $x_1=4$, $y_1=3$: $$y - 3 = 2(x - 4)$$ Simplify: $$y - 3 = 2x - 8$$ $$y = 2x - 8 + 3$$ $$y = 2x - 5$$ --- 2. **Find the equation of the line joining the points $(-2,1)$ and $(3,2)$.** First, find the gradient $m$: $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 1}{3 - (-2)} = \frac{1}{5}$$ Use point-slope form with point $(-2,1)$: $$y - 1 = \frac{1}{5}(x + 2)$$ Simplify: $$y - 1 = \frac{1}{5}x + \frac{2}{5}$$ $$y = \frac{1}{5}x + \frac{2}{5} + 1$$ $$y = \frac{1}{5}x + \frac{7}{5}$$ --- 3. **Are the two lines $y = \frac{3}{4}x - 5$ and $4y - 3x + 10 = 0$ parallel?** Find gradient of second line: $$4y - 3x + 10 = 0 \Rightarrow 4y = 3x - 10 \Rightarrow y = \frac{3}{4}x - \frac{10}{4}$$ Gradient of second line is $\frac{3}{4}$. Gradient of first line is $\frac{3}{4}$. Since $m_1 = m_2 = \frac{3}{4}$, the lines are parallel. --- 4. **Find the gradient of a line perpendicular to the line $x + 3y - 4 = 0$.** Rewrite line in slope-intercept form: $$x + 3y - 4 = 0 \Rightarrow 3y = -x + 4 \Rightarrow y = -\frac{1}{3}x + \frac{4}{3}$$ Gradient of given line is $m_1 = -\frac{1}{3}$. Gradient of perpendicular line $m_2$ satisfies: $$m_1 \times m_2 = -1$$ So: $$-\frac{1}{3} \times m_2 = -1 \Rightarrow m_2 = 3$$ --- 5. **Find the equation of the line that passes through the point $(6,-2)$ and is parallel to the line $2x - 3y + 4 = 0$.** Rewrite given line in slope-intercept form: $$2x - 3y + 4 = 0 \Rightarrow -3y = -2x - 4 \Rightarrow y = \frac{2}{3}x + \frac{4}{3}$$ Gradient of given line is $m = \frac{2}{3}$. Equation of line parallel to this and passing through $(6,-2)$: $$y - y_1 = m(x - x_1)$$ $$y + 2 = \frac{2}{3}(x - 6)$$ Simplify: $$y + 2 = \frac{2}{3}x - 4$$ $$y = \frac{2}{3}x - 4 - 2$$ $$y = \frac{2}{3}x - 6$$