Letโ€™s suppose we need to find

equation of line passing through

(โˆ’1, 3), (3, โˆ’5)

ย 

If we try to draw the graph, we see that

13.jpg

Now,

Letโ€™s take any point (x, y) on the line

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

So, their Area of triangle = 0

Area of triangle = 0

Finding equation of line using Determinants - Part 2

ย 

x (3 ร— 1 โˆ’ (โˆ’5) ร— 1) โˆ’ y ((โˆ’1) ร— 1 โˆ’ 3 ร— 1) + 1 ((โˆ’1) ร— (โˆ’5) โˆ’ 3 ร— 3) = 0

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

8x + 4y โˆ’ 4 = 0

4 (2x + y โˆ’ 1) = 0

2x + y โˆ’ 1 = 0

ย 

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

ย 

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

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

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

ย 

Therefore,

Area of triangle formed by them = 0

Finding equation of line using Determinants - Part 3

๐‘ฅ ((โˆ’1) ร— 1 โˆ’ 1) โˆ’ y (1 ร— 1 โˆ’ 4 ร— 1) + 1 (1 ร— 1 โˆ’4 ร— (โˆ’1)) = 0

๐‘ฅ (โˆ’1 โˆ’ 1) โˆ’ y (1 โˆ’ 4) + 1 (1 + 4) = 0

โˆ’2๐‘ฅ + 3y + 5 = 0

3y โˆ’ 2๐‘ฅ + 5 = 0

Thus, the required condition of the line is 3y โˆ’ 2๐’™ + 5 = 0

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Transcript

|โ– 8(๐‘ฅ_1&๐‘ฆ_1&1@๐‘ฅ_2&๐‘ฆ_2&1@๐‘ฅ_3&๐‘ฆ_3&1)| = 0 |โ– 8(๐‘ฅ&๐‘ฆ&1@โˆ’1&3&1@3&โˆ’5&1)| = 0 |โ– 8(๐‘ฅ_1&๐‘ฆ_1&1@๐‘ฅ_2&๐‘ฆ_2&1@๐‘ฅ_3&๐‘ฆ_3&1)| = 0 |โ– 8(๐‘ฅ&๐‘ฆ&1@1&โˆ’1&1@4&1&1)| = 0

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