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