English

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

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Question

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

Sum
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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).

Again, area of triangle = `1/2 |(x_1,y_1,1),(x_2,y_2,1),(x_3,y_3,1)|`

x1 = x, y1 = y, x2 = 1, y2 = 2, x3 = 3, y3 = 6

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

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

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

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

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

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

= −2x + y

The points are collinear.

Therefore the area of ​​the triangle 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 Part 1 and 2 [English] Class 12
Chapter 4 Determinants
Exercise 4.3 | Q 4.1 | Page 123
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