Thread: 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 x-axis.


2.
find the angle between the straight line 3x-2y=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 x-axis.

Alright, they gave you the angle (which is in degrees, not celsius) between the straight line and the positive direction of the x-axis. Recall that the slope, m, is:


m = \frac{\text{change\ in\ y}}{\text{change\ in\ x}} = \frac{y_2-y_1}{x_2-x_1}

This looks very similar to the tangent ratio, which is:

\tan{\theta} = \frac{y}{x}

Think of the straight line, change in y, and change in x as a triangle. Thus:


m = \tan{\theta} = \tan{30^{o}} = \frac{\sqrt{3}}{3}

Substitute in point-slope form equation:

y - y_1 = m(x - x_1)

2. find the angle between the straight line 3x-2y=4 and the line joining the points (-2,-1) and (4,1)

First, rearrange the given equation:

3x - 2y = 4

2y = 3x - 4

y = \frac{3}{2}x - 2

Thus, the slope of this equation (we call it m_1) is:

m_1 = \frac{3}{2}

Now, find the slope of the line that contains the two given points (we'll call it m_2)


m_2 = \frac{1 - (-1)}{4-(-2)} = \frac{2}{6} = \frac{1}{3}

Recall that:

\tan{\theta} = m

So, find the angle of both slopes and subtract them:


\tan^{-1}{m_1} - \tan^{-1}{m_2}

And voila, you got your angle