The straight line $L_1$ passes through the points $(-1, 3)$ and $(11, 12)$. (a) Find an equation for $L_1$, in the form $ax + by + c = 0$, where $a, b$ and $c$ are integers. The line $L_2$ has equation $3y + 4x - 30 = 0$. (b) Find the coordinates of the point of intersection of $L_1$ and $L_2$.

A-Level Edexcel Maths Pure Transformations of Functions: The straight line $L

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

Question 8

The straight line $L_1$ passes through the points $(-1, 3)$ and $(11, 12)$. (a) Find an equation for $L_1$, in the form $ax + by + c = 0$, where $a, b$ and $c$ ... show full transcript

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

Step 1

Find an equation for $L_1$, in the form $ax + by + c = 0$

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Answer

To find the equation of the line $L_1$ that passes through the points $(-1, 3)$ and $(11, 12)$ , we first calculate the slope $m$ of the line.

The slope $m$ can be calculated using the formula:

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

Substituting the coordinates:

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

Now, using the point-slope form of the line equation, which is given by:

$y - y_1 = m(x - x_1)$

We can use the point $(-1, 3)$ :

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

Expanding this gives:

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

Rearranging to get $ax + by + c = 0$ form:

$y - \frac{3}{4}x - \frac{3}{4} + 3 = 0$

This simplifies to:

$-\frac{3}{4}x + y + \frac{9}{4} = 0$

Multiplying through by 4 to eliminate the fraction yields:

$-3x + 4y + 9 = 0$

Thus, the equation of line $L_1$ is:

$3x - 4y - 9 = 0$

Step 2

Find the coordinates of the point of intersection of $L_1$ and $L_2$

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Answer

To find the intersection point of the lines $L_1$ and $L_2$ , we need to solve the system of equations formed by their equations:

From equation of $L_1$ :
$3x - 4y - 9 = 0$
Rearranging gives us:
$4y = 3x - 9$
$y = \frac{3}{4}x - \frac{9}{4}$
From the equation of $L_2$ :
$3y + 4x - 30 = 0$
Rearranging gives us:
$3y = -4x + 30$
$y = -\frac{4}{3}x + 10$

Now, we can set the two expressions for $y$ equal to each other:

$\frac{3}{4}x - \frac{9}{4} = -\frac{4}{3}x + 10$

To eliminate the fractions, we can multiply through by 12 (the least common multiple of the denominators):

$12 \left( \frac{3}{4}x \right) - 12 \left( \frac{9}{4} \right) = 12 \left( -\frac{4}{3}x \right) + 12 \times 10$

This simplifies to:

$9x - 27 = -16x + 120$

Now, solving for $x$ :

$9x + 16x = 120 + 27$ $25x = 147$ $x = \frac{147}{25}$

Now substituting back to find $y$ using one of the original equations, let's use $L_2$ :

$y = -\frac{4}{3}\left( \frac{147}{25} \right) + 10$

Calculating this yields:

$y = -\frac{588}{75} + \frac{750}{75}$ $y = \frac{162}{75}$

Thus, the coordinates of the point of intersection of lines $L_1$ and $L_2$ are:

$\left( \frac{147}{25}, \frac{162}{75} \right)$

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

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

Find an equation for $L_1$, in the form $ax + by + c = 0$

Find the coordinates of the point of intersection of $L_1$ and $L_2$

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