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
**

Are ads bothering you?