Suppose we have Line 1 as y=2x-20 and Line 2 as y= 3x-32. The point of intersection of these two lines is (4,-12). Now my question is how do you get the angle at which these two lines meet both the x and y axis and themselves?
Suppose we have Line 1 as y=2x-20 and Line 2 as y= 3x-32. The point of intersection of these two lines is (4,-12). Now my question is how do you get the angle at which these two lines meet both the x and y axis and themselves?
1. Let $\displaystyle \alpha$ denote the angle between the x-axis and the line. Then the slope of the line is
$\displaystyle m = \tan(\alpha)$
2. Draw a parallel to the x-axis through the point of intersection. You'll easily find the angle $\displaystyle \alpha$ at this point again.
3. The angle between 2 lines is the difference of angles $\displaystyle \alpha$
By the way: With your equations of the 2 lines the point of intersection is at (12, 4)
Ok now I know how to find the angle between the x-axis and the lines simply by substituting for the value of m and getting the arctan. What of the y axis? Also for number 3, you know each line can have an angle betwen itself and the x axis or the y axis so when you said that the angle between 2 lines is the difference of angles, do you mean the difference between the alpha angle for y and the line or the difference between the angle alpha of x and the line? I understand that there are four angles.
Hi Keep,
These two lines intersect at (12, 4), not (4, -12).
You can find the angle each line makes with the x-axis by taking the inverse tangent of their slopes.
$\displaystyle y=2x-20$ makes an angle of $\displaystyle \tan^{-1}(2)\approx 63.4^{\circ}$ with the x-axis.
$\displaystyle y=3x-32$ makes an angle of $\displaystyle \tan^{-1}(3)\approx 71.6^{\circ}$ with the x-axis.
To find the angle the lines make with themselves, recognize that a triangle is formed with their intersection point and their x-intercepts. The exterior angle is 71.6 degrees. That is equal to the sum of the two remote interior angles. So the acute angle at the point of intersection of the two lines is 71.6 - 63.4 = 8.2 degrees.
Use the x-intercepts and y-intercepts and the origin to help you determine the angle each line makes with the y-axis.
So we have the y intercept which is -20 for L1 and -32 for lines. Their x intercept are 10 and 10.67 respectively. So I divide the x intercept by y intercept to get tan alpha. So alpha for L1 would be arctan -0.5 and the other as -0.3? But the calculator apparently calculates for a positive arctan.
The angles the lines make with the y-axis will be the complement of the angles they made with the x-axis.
Line 1 made a 63.4 degree angle with the x-axis. It will make a 26.6 degree angle with the y-axis.
Line 2 made a 71.6 degree angle with the x-axis. It will make a 18.4 degree angle with the y-axis.