Let’s suppose we need to find

equation of line passing through

(−1, 3), (3, −5)

 

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), (−1, 3), (3, −5) are on the same line

So, their Area of triangle = 0

Area of triangle = 0

14.jpg

 

x (3 × 1 − (−5) × 1) − y ((−1) × 1 − 3 × 1) + 1 ((−1) × (−5) − 3 × 3) = 0

x (3 + 5) − y (−1 − 3) + (5 − 9) = 0

8x + 4y − 4 = 0

4 (2x + y − 1) = 0

2x + y − 1 = 0

 

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

 

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

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

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

 

Therefore,

Area of triangle formed by them = 0

15.jpg

๐‘ฅ ((−1) × 1 − 1) − y (1 × 1 − 4 × 1) + 1 (1 × 1 −4 × (−1)) = 0

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

−2๐‘ฅ + 3y + 5 = 0

3y − 2๐‘ฅ + 5 = 0

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

  1. Chapter 4 Class 12 Determinants
  2. Concept wise

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

About the Author

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.