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

Finding equation of line using Determinants - Part 2

 

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

Finding equation of line using Determinants - Part 3

𝑥 ((−1) × 1 − 1) − y (1 × 1 − 4 × 1) + 1 (1 × 1 −4 × (−1)) = 0

𝑥 (−1 − 1) − y (1 − 4) + 1 (1 + 4) = 0

−2𝑥 + 3y + 5 = 0

3y − 2𝑥 + 5 = 0

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

Go Ad-free

Transcript

|■8(𝑥_1&𝑦_1&1@𝑥_2&𝑦_2&1@𝑥_3&𝑦_3&1)| = 0 |■8(𝑥&𝑦&1@−1&3&1@3&−5&1)| = 0 |■8(𝑥_1&𝑦_1&1@𝑥_2&𝑦_2&1@𝑥_3&𝑦_3&1)| = 0 |■8(𝑥&𝑦&1@1&−1&1@4&1&1)| = 0

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo