Given a straight line with equation coefficients as a, b & c(ax + by + c = 0), the task is to find the area of the triangle formed by the axes of co-ordinates and this straight line.
Input: a = -2, b = 4, c = 3 Output: 0.5625 Input: a = 4, b = 3, c = 12 Output: 6
- Let PQ be the straight line having AB, the line segment between the axes.
The equation is,
ax + by + c = 0
- so, in intercept form it can be expressed as,
x/(-c/a) + y/(-c/b) = 1
- So, the x-intercept = -c/a
the y-intercept = -c/b
- So, it is very clear now the base of the triangle AOB will be -c/a
and the base of the triangle AOB will be -c/b
- So, area of the triangle
Below is the implementation of the above approach:
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