Letβs suppose we need to find
equation of line passing through
(β1, 3), (3, β5)
Β
If we try to draw the graph, we see that
Now,
Letβs take any point (x, y) on the line
We see that (x, y), (β1, 3), (3, β5) are on the same line
So, their Area of triangle = 0
Area of triangle = 0
Β
x (3 Γ 1 β (β5) Γ 1) β y ((β1) Γ 1 β 3 Γ 1) + 1 ((β1) Γ (β5) β 3 Γ 3) = 0
x (3 + 5) β y (β1 β 3) + (5 β 9) = 0
8x + 4y β 4 = 0
4 (2x + y β 1) = 0
2x + y β 1 = 0
Β
So, equation of line is 2 x + y β 1 = 0
Β
Find equation of line passing through (1, β1) & (4, 1), using determinants
Let (x, y) be a point on the required line
So, (x, y), (1, β1) & (4, 1) are in a same line
Β
Therefore,
Area of triangle formed by them = 0
π₯ ((β1) Γ 1 β 1) β y (1 Γ 1 β 4 Γ 1) + 1 (1 Γ 1 β4 Γ (β1)) = 0
π₯ (β1 β 1) β y (1 β 4) + 1 (1 + 4) = 0
β2π₯ + 3y + 5 = 0
3y β 2π₯ + 5 = 0
Thus, the required condition of the line is 3y β 2π + 5 = 0
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