Given two points P and Q in the coordinate plane, find the equation of the line passing through both the points.
Input : P(3, 2) Q(2, 6) Output : 4x + 1y = 14 Input : P(0, 1) Q(2, 4) Output : 3x + -2y = -2
Let the given two points be P(x1, y1) and Q(x2, y2). Now, we find the equation of line formed by these points.
Any line can be represented as,
ax + by = c
Let the two points satisfy the given line. So, we have,
ax1 + by1 = c
ax2 + by2 = c
We can set the following values so that all the equations hold true,
a = y2 - y1 b = x1 - x2 c = ax1 + by1
These can be derived by first getting the slope directly and then finding the intercept of the line. OR these can also be derived cleverly by a simple observation as under:
ax1 + by1 = c ...(i) ax2 + by2 = c ...(ii) Equating (i) and (ii), ax1 + by1 = ax2 + by2 => a(x1 - x2) = b(y2 - y1) Thus, for equating LHS and RHS, we can simply have, a = (y2 - y1) AND b = (x1 - x2) so that we have, (y2 - y1)(x1 - x2) = (x1 - x2)(y2 - y1) AND Putting these values in (i), we get, c = ax1 + by1
Thus, we now have the values of a, b and c which means that we have the line in the coordinate plane.
The line passing through points P and Q is: 4x + 1y = 14
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Improved By : ash264