Let’s suppose we need to find
equation of line passing through
(−1, 3), (3, −5)
If we try to draw the graph, we see that
Now,
Let’s take any point (x, y) on the line
We see that (x, y), (−1, 3), (3, −5) are on the same line
So, their Area of triangle = 0
Area of triangle = 0
x (3 × 1 − (−5) × 1) − y ((−1) × 1 − 3 × 1) + 1 ((−1) × (−5) − 3 × 3) = 0
x (3 + 5) − y (−1 − 3) + (5 − 9) = 0
8x + 4y − 4 = 0
4 (2x + y − 1) = 0
2x + y − 1 = 0
So, equation of line is 2 x + y − 1 = 0
Find equation of line passing through (1, −1) & (4, 1), using determinants
Let (x, y) be a point on the required line
So, (x, y), (1, −1) & (4, 1) are in a same line
Therefore,
Area of triangle formed by them = 0
๐ฅ ((−1) × 1 − 1) − y (1 × 1 − 4 × 1) + 1 (1 × 1 −4 × (−1)) = 0
๐ฅ (−1 − 1) − y (1 − 4) + 1 (1 + 4) = 0
−2๐ฅ + 3y + 5 = 0
3y − 2๐ฅ + 5 = 0
Thus, the required condition of the line is 3y − 2๐ + 5 = 0