Question

Show that the origin is with in the triangle whose angular points are (2, 1),

(3, 2) and (-4, -1).




The vertices of the ∆ l e ABC are A = (2, 1), = (3, -2), = (-4, -1)

The equation of BC is L x 7y 11 0 =+ + = ---(1)

The equation of CA is ⇒ L' x 3y 1 0 = − += ---(2)

Equation of AB is L” = - 3x + y – 7 = 0 ---(3)

L” (-4, -1) = 3 (-4) -1 -7 = - 20 is negative.

L” (0, 0) = 3(0) + 0 – 7 = - 7 is negative

So (-4, -1), (0. 0) lie on the same side of AB

Hence O (0, 0) lies to the left of AB ---(4)

L’ (3, -2) = 3 – 3(-2) + 1 = 10 is positive

L’ (0, 0) =0 – 3(0) + 1 = 1 is positive

So (0, 0), (3, -2) lie on the same side of AC both down of AC ---(5)

L (2, 1) = 2 + 7 (1) + 11 = 20 is positive

L (0, 0) = 0 + 7 (0) + 11 = 11 is positive

So (0, 0) and (2, 1) lie on the same side of BC.

So (0, 0) lie upwards form BC ---(6)

From (4), (5), (6), it follows O (0, 0) lies down-wards of AC, upwards of BC, and to

the left of AB. So O (0, 0) will lie inside the ∆ ABC.​

Answer 1

Answer:

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