One of the angles between two non-parallel lines is always less than [tex]90^o[/tex]; i.e. it is acute. The measure of the acute angle between L1 and L2 is [tex]52^o[/tex]
Start by calculating the point of intersection between [tex]2y = x - 13[/tex] and [tex]3y + x + 12 = 0[/tex]
Make x the subject in [tex]2y = x - 13[/tex]
[tex]x = 2y +13[/tex]
Substitute [tex]x = 2y +13[/tex] in [tex]3y+x+12=0[/tex]
[tex]3y+2y+13+12=0[/tex]
[tex]5y + 25 = 0[/tex]
[tex]5y = -25[/tex]
Divide both sides by 5
[tex]y=-5[/tex]
Recall that: [tex]x = 2y +13[/tex]
[tex]x = 2 \times -5 + 13[/tex]
[tex]x = 3[/tex]
So, the point of intersection is (3,-5)
Next, calculate the slopes.
L1 passes through (-4,7).
So, the slope (m) is:
[tex]m = \frac{y_2 - y_1}{x_2 - x_1}[/tex]
[tex]m = \frac{7--5}{-4-3}[/tex]
[tex]m = \frac{12}{-7}[/tex]
[tex]m_1 = -\frac{12}{7}[/tex]
Also, we have:
[tex]2x - 5y = 4[/tex]
Make y the subject
[tex]5y = 2x - 4[/tex]
[tex]y = \frac 25x - \frac 45[/tex]
A linear equation is represented as: [tex]y = mx + b\\[/tex]
Where: [tex]m \to slope[/tex]
So, we have:
[tex]m =\frac 25[/tex]
Because L2 is perpendicular to [tex]2x - 5y = 4[/tex], the slope of L2 is
[tex]m_2 = \frac 1m[/tex]
[tex]m_2 = \frac 1{2/5}[/tex]
[tex]m_2 = \frac 52[/tex]
So, we have:
[tex]m_1 = -\frac{12}{7}[/tex] --- slope of L1
[tex]m_2 = \frac 52[/tex] ----- slope of L2
The angle between them is then calculated as follows:
[tex]\tan(\theta) = \frac{m_1 - m_2}{1 + m_1 \times m_2}[/tex]
So, we have:
[tex]\tan(\theta) = \frac{-12/7 - 5/2}{1 -12/7 \times 5/2}[/tex]
[tex]\tan(\theta) = \frac{-4.2143}{-3.2857}[/tex]
[tex]\tan(\theta) = 1.2826[/tex]
Take arc tan of both sides
[tex]\theta = \tan^{-1}(1.2826)[/tex]
[tex]\theta = 52^o[/tex] --- Approximately
Hence, the acute angle between L1 and L2 is [tex]52^o[/tex]
Read more about acute angles between lines at:
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