
Linear Lines
1.
Find the equation of a straight line which passes through the point (2,3) and is inclined at 30 Celsius to the positive direction of the xaxis.
2.
find the angle between the straight line 3x2y=4 and the line joining the points (2,1) and (4,1)
please guide me step by step
thanks

1. Find the equation of a straight line which passes through the point (2,3) and is inclined at 30 Celsius to the positive direction of the xaxis.
Alright, they gave you the angle (which is in degrees, not celsius) between the straight line and the positive direction of the xaxis. Recall that the slope, m, is:
http://www.gnux.be/latex/data/7f7b7c...cbbf75081f.png
m = \frac{\text{change\ in\ y}}{\text{change\ in\ x}} = \frac{y_2y_1}{x_2x_1}
This looks very similar to the tangent ratio, which is:
http://www.gnux.be/latex/data/4eb4e7...8131619926.png
\tan{\theta} = \frac{y}{x}
Think of the straight line, change in y, and change in x as a triangle. Thus:
http://www.gnux.be/latex/data/b3b44c...65ab7bce5f.png
m = \tan{\theta} = \tan{30^{o}} = \frac{\sqrt{3}}{3}
Substitute in pointslope form equation:
http://www.gnux.be/latex/data/c53be4...0af2a2911b.png
y  y_1 = m(x  x_1)
2. find the angle between the straight line 3x2y=4 and the line joining the points (2,1) and (4,1)
First, rearrange the given equation:
3x  2y = 4
2y = 3x  4
http://www.gnux.be/latex/data/b9984c...a6e085759e.png
y = \frac{3}{2}x  2
Thus, the slope of this equation (we call it m_1) is:
http://www.gnux.be/latex/data/c3e744...dfe36d1925.png
m_1 = \frac{3}{2}
Now, find the slope of the line that contains the two given points (we'll call it m_2)
http://www.gnux.be/latex/data/611b70...9c3500f70f.png
m_2 = \frac{1  (1)}{4(2)} = \frac{2}{6} = \frac{1}{3}
Recall that:
http://www.gnux.be/latex/data/6208c5...e6a0f37146.png
\tan{\theta} = m
So, find the angle of both slopes and subtract them:
http://www.gnux.be/latex/data/a615ce...0f698d9105.png
\tan^{1}{m_1}  \tan^{1}{m_2}
And voila, you got your angle :D