Find equation of line joining (1, 2) and (3, 6) using the determinant. - Mathematics

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Sum

Find equation of line joining (1, 2) and (3, 6) using the determinant.

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Solution

Let there be a point (x, y).

Therefore the vertices of the triangle will be (x, y), (1, 2), (3,6).

`Delta` area of `Delta` = `1/2 abs ((x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1))`

`x_1 = x, y_1 = y, x_2 = 1, y_2 = 2, x_3 = 3, y_3 = 6`

`= 1/2 abs ((x,y,1),(1,2,1),(3,6,1))`

`= 1/2 [x (2 - 6) - y (1 - 3) + 1(6 - 6)]`

`= 1/2 [x xx (-4) - y (-2) + 1 xx 0]`

`= 1/2 [- 4x + 2y]`

`= 1/2 xx 2 (-2x + y)`

`= -2x + y`

The points are collinear.

So, the area of ​​the triangle is

Therefore the area of ​​`Delta` will be zero.

`=> 0 = -2x + y`

`=> 2x - y = 0`

= y = 2x

This is the required equation.

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Chapter 4: Determinants - Exercise 4.3 [Page 123]

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NCERT Mathematics Class 12
Chapter 4 Determinants
Exercise 4.3 | Q 4.1 | Page 123

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