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Let’s suppose we need to find

equation of line passing through

(βˆ’1, 3), (3, βˆ’5)

Β 

If we try to draw the graph, we see that

13.jpg

Now,

Let’s take any point (x, y) on the line

We see that (x, y), (βˆ’1, 3), (3, βˆ’5) are on the same line

So, their Area of triangle = 0

Area of triangle = 0

Finding equation of line using Determinants - Part 2

Β 

x (3 Γ— 1 βˆ’ (βˆ’5) Γ— 1) βˆ’ y ((βˆ’1) Γ— 1 βˆ’ 3 Γ— 1) + 1 ((βˆ’1) Γ— (βˆ’5) βˆ’ 3 Γ— 3) = 0

x (3 + 5) βˆ’ y (βˆ’1 βˆ’ 3) + (5 βˆ’ 9) = 0

8x + 4y βˆ’ 4 = 0

4 (2x + y βˆ’ 1) = 0

2x + y βˆ’ 1 = 0

Β 

So, equation of line is 2 x + y βˆ’ 1 = 0

Β 

Find equation of line passing through (1, βˆ’1) & (4, 1), using determinants

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

So, (x, y), (1, βˆ’1) & (4, 1) are in a same line

Β 

Therefore,

Area of triangle formed by them = 0

Finding equation of line using Determinants - Part 3

π‘₯ ((βˆ’1) Γ— 1 βˆ’ 1) βˆ’ y (1 Γ— 1 βˆ’ 4 Γ— 1) + 1 (1 Γ— 1 βˆ’4 Γ— (βˆ’1)) = 0

π‘₯ (βˆ’1 βˆ’ 1) βˆ’ y (1 βˆ’ 4) + 1 (1 + 4) = 0

βˆ’2π‘₯ + 3y + 5 = 0

3y βˆ’ 2π‘₯ + 5 = 0

Thus, the required condition of the line is 3y βˆ’ 2𝒙 + 5 = 0

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Transcript

|β– 8(π‘₯_1&𝑦_1&1@π‘₯_2&𝑦_2&1@π‘₯_3&𝑦_3&1)| = 0 |β– 8(π‘₯&𝑦&1@βˆ’1&3&1@3&βˆ’5&1)| = 0 |β– 8(π‘₯_1&𝑦_1&1@π‘₯_2&𝑦_2&1@π‘₯_3&𝑦_3&1)| = 0 |β– 8(π‘₯&𝑦&1@1&βˆ’1&1@4&1&1)| = 0

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.