Area of a Δ ABC when its three vertices are given.
If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle then its area is given by
Area of ΔABC = 1 /2 [x1(y2 - y3) + x2( y3 - y1)+ x3(y1 - y2)] square units.
If the area of triangle is zero then the points are called collinear points.
If three points A(x1, y1),B(x2, y2) and C(x3, y3) are collinear then
Area of triangle =0, [x1(y2 - y3) + x2( y3- y1)+ x3(y1 - y2)] = 0.
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Solution:
i) when A is taken as origin , AD & AB as coordinates axes i.e X & Y axes. Coordinates of P , Q & R are (4,6) , (3,2), (6,5)
ii) i) when C is taken as origin & CB & CD as coordinates axes i.e X & Y axes. Coordinates of P , Q & R are (12,2) , (13,6), (10,3)
Area of Δ according to both cases are as follows
Case I:
Area of ΔPQR in case of origin A & AD & AB as coordinate axes.
= ½ [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]
= ½ ( - 12 – 3 + 24
Area of ΔPQR= 9/2 sq unit
Case II
Area of Δ PQR in case of origin C & CB & CD as coordinate axes.
= ½ [x1(y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)]
= ½ [ -12( - 6 + 3) + - 13 (- 3 + 2) + - 10( - 2 + 6)]
= ½ ( 36 + 13 – 40)
Area of Δ PQR= 9/2 sq unit
Hence, we observe that area of Δ is same in both case because triangle remains the same no matter which point is considered as origin.----------------------------------------------------------------------------------------------------Hope this will help you...
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Solution: