ax + by + c = 0 exmple
Answers
claim that an equation of the form Ax + By + C = 0, will represent a straight line. Here A, B, C are arbitrary constants (A and B cannot be both 0), and x and y are variables (which represent the coordinates of points on the line).
Here’s a proof (somewhat).
Suppose B ≠ 0, then on dividing the equation by B and rearranging the terms we get
y = (-A/B)x + (-C/B)
By putting -A/B = m and -C/B = c, the above equation becomes
y = mx + c
This looks familiar.
This is a line of slope m, and y-intercept c, as we derived here. Therefore, the equation Ax + By + C = 0 represents a line with slope -A/B and y-intercept -C/B.
In case B = 0, then the equation Ax + By + C = 0 will become
x = -C/A
This represents a line parallel to the Y axis (explained here).
Similarly, if A = 0, then the equation Ax + By + C = 0 will become
y = -C/B
This represents a line parallel to the X axis (explained here).
In each case, the equation Ax + By + C = 0 represented a straight line.
But what if we hadn’t derived these equations previously? I’ll give one more proof.
The idea is to prove that any three points taken on the curve represented by Ax + By + C = 0 are collinear. Because if they are, then that curve can be nothing else but a line!
Let P(x1, y1) and Q(x2, y2) and R(x3, y3) be any three points on the curve whose equation is Ax + By + C = 0.
Then the coordinates of these three points must satisfy the equation. We get the following three relations.
Ax1 + By1 + C = 0 … I
Ax2 + By2 + C = 0 … II
Ax3 + By3 + C = 0 … III
Now, the aim is to prove that, if these relations hold true, then the points must be collinear. Let’s try to prove that.
Subtracting I from II, we get
A(x2 – x1) = B(y2 – y1)
(y2 – y1)/(x2 – x1) = A/B … IV
Similarly, after subtracting II from III, we get
(y3 – y2)/(x3 – x2) = A/B … V
From IV and V, we’ll get
(y2 – y1)/(x2 – x1) = (y3 – y2)/(x3 – x2)
This, on rearranging, gives
x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) = 0
Finally, on dividing this by 2, we get
1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)] = 0
This means that the area of the triangle PQR is 0, proving that the points are collinear. (We’ve done something like this previously.)