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Exa1- Find the area of the triangle whose vertices are (1, 2), (2, 1) and (0, 0).
Sol. By the formula, area of the triangle

[x 1 (y 2 - y 3 ) + x 2 (y 3 - y 1 ) + x 3 (y 1 - y 2 )]

1  3
4
[1 (1 - 0) + 2(0 - 2) + 0(2 - 1)]= [ 1  ] 
2 2

Rejecting the negative sign, as area can't be negative we get the required area as
square units.

Exa2 Show that the points (2, 4), (0, 1) and (4, 7) are collinear.

Sol. Area of the triangle formed by three points is given by

Substituting the values, we get,

= [2(1 - 7) + 0(7 - 4) + 4(4 - 1)]

[- 12 + 12] = 0
Hence, the points are collinear.
Exa3 Find the condition that the point (x, y) may lie on the line joining (2, 3) and (-5, -6).
Sol. The point (x, y) lies on the line joining (2, 3) and (-5, -6)
(x, y), (2, 3), (-5, -6) are collinear
So, the area of triangle formed by them is zero i.e.
1/2[x(3 + 6) + 2(-6 - y) -5(y - 3)] = 0
9x - 7y + 3 = 0

Exa4 Show that the following triple points are collinear (2, 3), (0, 1), (4, 5).

Sol. Area of triangle formed by given points is
1/2[2(1 - 5) + 0(5 - 3) + 4(3 - 1)]
= 1/2 [-8+ 0 + 8]
= 0
Since the area of the triangle is zero, so the points are collinear

Exa5 For what value of x will the points (x, -1), (2, 1) and (4, 5) lie on a line?

Sol. Area of triangle formed by the given points

[x (1 -5) + 2(5+1) + 4(-1 - 1)]

[-4x +12 -8]

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