Question

Two lines L1 and L2 pass through

Answer 1

One of the angles between two non-parallel lines is always less than $90^o$; i.e. it is acute. The measure of the acute angle between L1 and L2 is $52^o$

Start by calculating the point of intersection between $2y = x - 13$ and $3y + x + 12 = 0$

Make x the subject in $2y = x - 13$

$x = 2y +13$

Substitute $x = 2y +13$ in $3y+x+12=0$

$3y+2y+13+12=0$

$5y + 25 = 0$

$5y = -25$

Divide both sides by 5

$y=-5$

Recall that: $x = 2y +13$

$x = 2 \times -5 + 13$

$x = 3$

So, the point of intersection is (3,-5)

Next, calculate the slopes.

L1 passes through (-4,7).

So, the slope (m) is:

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

$m = \frac{7--5}{-4-3}$

$m = \frac{12}{-7}$

$m_1 = -\frac{12}{7}$

Also, we have:

$2x - 5y = 4$

Make y the subject

$5y = 2x - 4$

$y = \frac 25x - \frac 45$

A linear equation is represented as: $y = mx + b$

Where: $m \to slope$

So, we have:

$m =\frac 25$

Because L2 is perpendicular to $2x - 5y = 4$, the slope of L2 is

$m_2 = \frac 1m$

$m_2 = \frac 1{2/5}$

$m_2 = \frac 52$

So, we have:

$m_1 = -\frac{12}{7}$ --- slope of L1

$m_2 = \frac 52$ ----- slope of L2

The angle between them is then calculated as follows:

$\tan(\theta) = \frac{m_1 - m_2}{1 + m_1 \times m_2}$

So, we have:

$\tan(\theta) = \frac{-12/7 - 5/2}{1 -12/7 \times 5/2}$

$\tan(\theta) = \frac{-4.2143}{-3.2857}$

$\tan(\theta) = 1.2826$

Take arc tan of both sides

$\theta = \tan^{-1}(1.2826)$

$\theta = 52^o$ --- Approximately

Hence, the acute angle between L1 and L2 is $52^o$

Read more about acute angles between lines at:

brainly.com/question/67538