Equation of line using determinant

Chapter 4 Class 12 Determinants
Concept wise

Letβs suppose we need to find

equation of line passing through

(β1, 3), (3, β5)

Β

If we try to draw the graph, we see that

Now,

Letβs take any point (x, y) on the line

We see that (x, y), (β1, 3), (3, β5) are on the same line

So, their Area of triangle = 0

Area of triangle = 0

Β

x (3 Γ 1 β (β5) Γ 1) β y ((β1) Γ 1 β 3 Γ 1) + 1 ((β1) Γ (β5) β 3 Γ 3) = 0

x (3 + 5) β y (β1 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, β1) & (4, 1), using determinants

Let (x, y) be a point on the required line

So, (x, y), (1, β1) & (4, 1) are in a same line

Β

Therefore,

Area of triangle formed by them = 0

π₯ ((β1) Γ 1 β 1) β y (1 Γ 1 β 4 Γ 1) + 1 (1 Γ 1 β4 Γ (β1)) = 0

π₯ (β1 β 1) β y (1 β 4) + 1 (1 + 4) = 0

β2π₯ + 3y + 5 = 0

3y β 2π₯ + 5 = 0

Thus, the required condition of the line is 3y β 2π + 5 = 0

Get live Maths 1-on-1 Classs - Class 6 to 12

Transcript

|β 8(π₯_1&π¦_1&[email protected]π₯_2&π¦_2&[email protected]π₯_3&π¦_3&1)| = 0 |β 8(π₯&π¦&[email protected]β1&3&[email protected]&β5&1)| = 0 |β 8(π₯_1&π¦_1&[email protected]π₯_2&π¦_2&[email protected]π₯_3&π¦_3&1)| = 0 |β 8(π₯&π¦&[email protected]&β1&[email protected]&1&1)| = 0