The straight line $L_1$ passes through the points $(-1, 3)$ and $(11, 12)$ - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 1

To find the equation of the line $L_1$ that passes through two given points, we first calculate the slope $m$ using the formula:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

For points $(-1, 3)$ (let this be $(x_1, y_1)$) and $(11, 12)$ (let this be $(x_2, y_2)$), we have:

$m = \frac{12 - 3}{11 - (-1)} = \frac{9}{12} = \frac{3}{4}$

Now using the point-slope form of the line, $y - y_1 = m(x - x_1)$, we substitute:

$y - 3 = \frac{3}{4}(x + 1)$

Expanding this gives:

\Rightarrow y = \frac{3}{4}x + \frac{3}{4} + 3 \Rightarrow y = \frac{3}{4}x + \frac{15}{4}$$ To convert this to the form $ax + by + c = 0$, we multiply through by 4: $$4y = 3x + 15 \Rightarrow -3x + 4y - 15 = 0$$ Thus, the equation for $L_1$ is: $$3x - 4y + 15 = 0.$$