3. Find the equation of the line through $$ (-2,3) $$ with slope (i) 4 (ii) -2 (iii) $$ \frac{3}{4} $$ (iv) $$ -\frac{2}{3} $$ 4. Find the equation of the line through $$ (0,0) $$ and whose slope is -3 . 5. Find the equation of the line through $$ (0,0) $$ and whose slope is (i) 3 (ii) -5 (iii) $$ \frac{1}{3} $$ (iv) $$ -\frac{3}{2} $$ What do you notice about the equation of each of these lines? 6. Find the slope of the line through $$ A(3,-4) $$ and $$ B(1,2) $$. Hence find the equation of the line $$ A B $$.

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3. For a line with slope $$ m $$ through a point $$(x_1, y_1)$$, the point‐slope form is  
$$
y - y_1 = m(x - x_1).
$$
Here, $$(x_1, y_1) = (-2,3)$$.

(i) For $$ m = 4 $$:  
$$
\begin{aligned}
y - 3 &= 4\,(x - (-2)) = 4(x+2),\[1mm]
y - 3 &= 4x + 8,\[1mm]
y &= 4x + 11.
\end{aligned}
$$

(ii) For $$ m = -2 $$:  
$$
\begin{aligned}
y - 3 &= -2\,(x+2),\[1mm]
y - 3 &= -2x - 4,\[1mm]
y &= -2x - 1.
\end{aligned}
$$

(iii) For $$ m = \frac{3}{4} $$:  
$$
\begin{aligned}
y - 3 &= \frac{3}{4}(x+2),\[1mm]
y - 3 &= \frac{3}{4}x + \frac{3}{2},\[1mm]
y &= \frac{3}{4}x + \frac{3}{2} + 3.
\end{aligned}
$$
Since $$ 3 = \frac{6}{2} $$,  
$$
y = \frac{3}{4}x + \frac{9}{2}.
$$

(iv) For $$ m = -\frac{2}{3} $$:  
$$
\begin{aligned}
y - 3 &= -\frac{2}{3}(x+2),\[1mm]
y - 3 &= -\frac{2}{3}x - \frac{4}{3},\[1mm]
y &= -\frac{2}{3}x - \frac{4}{3} + 3.
\end{aligned}
$$
Since $$ 3 = \frac{9}{3} $$,  
$$
y = -\frac{2}{3}x + \frac{5}{3}.
$$

4. For a line through $$(0,0)$$ with slope $$-3$$, the point‑slope form is:
$$
y - 0 = -3\,(x - 0),
$$
which simplifies to  
$$
y = -3x.
$$

5. For a line through $$(0,0)$$, the point‑slope form becomes  
$$
y = mx.
$$
Thus, for each case:

(i) For $$ m = 3 $$:  
$$
y = 3x.
$$

(ii) For $$ m = -5 $$:  
$$
y = -5x.
$$

(iii) For $$ m = \frac{1}{3} $$:  
$$
y = \frac{1}{3}x.
$$

(iv) For $$ m = -\frac{3}{2} $$:  
$$
y = -\frac{3}{2}x.
$$

Observation: Every line that goes through the origin $$(0,0)$$ has an equation of the form  
$$
y = mx,
$$
where $$ m $$ is the slope.

6. To find the slope of the line through $$ A(3,-4) $$ and $$ B(1,2) $$, use the slope formula  
$$
m = \frac{y_2 - y_1}{x_2 - x_1}.
$$
Let $$ A(3,-4) $$ be $$(x_1,y_1)$$ and $$ B(1,2) $$ be $$(x_2,y_2)$$:
$$
\begin{aligned}
m &= \frac{2 - (-4)}{1-3} = \frac{6}{-2} = -3.
\end{aligned}
$$

Using point-slope form with point $$ A(3,-4) $$:
$$
\begin{aligned}
y - (-4) &= -3(x - 3),\[1mm]
y + 4 &= -3x + 9,\[1mm]
y &= -3x + 5.
\end{aligned}
$$
3. (i) $$ y = 4x + 11 $$ (ii) $$ y = -2x - 1 $$ (iii) $$ y = \frac{3}{4}x + \frac{9}{2} $$ (iv) $$ y = -\frac{2}{3}x + \frac{5}{3} $$ 4. $$ y = -3x $$ 5. (i) $$ y = 3x $$ (ii) $$ y = -5x $$ (iii) $$ y = \frac{1}{3}x $$ (iv) $$ y = -\frac{3}{2}x $$ Observation: All lines through the origin have equations of the form $$ y = mx $$, where $$ m $$ is the slope. 6. Slope $$ m = -3 $$ Equation of line $$ AB $$: $$ y = -3x + 5 $$

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