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:
- Equation of straight line passing through a given point which bisects it into two equal line segments
- Represent a given set of points by the best possible straight line
- Length of the normal from origin on a straight line whose intercepts are given
- Find coordinates of the triangle given midpoint of each side
- Find minimum area of rectangle with given set of coordinates
- Area of Reuleaux Triangle
- Check if right triangle possible from given area and hypotenuse
- Program to find area of a triangle
- Area of Circumcircle of a Right Angled Triangle
- Area of Incircle of a Right Angled Triangle
- Area of a triangle inside a parallelogram
- Check if a triangle of positive area is possible with the given angles
- Find the altitude and area of an isosceles triangle
- Minimum height of a triangle with given base and area
- Area of circle which is inscribed in equilateral triangle
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