Area of Triangle Using Coordinate


Coordinate Geometry II - Concepts
Class - 10th CBSE Subjects
 
 
Concept Explanation
 

Area of Triangle Using Coordinate

We know that area of the trapezium ABCD whose two parallel sides are AB and DC and the distance between parallel sides is h, is given by

      large frac{1}{2}(AB+DC)h

Let ABC be a triangle with vertices large A(x_{1},y_{1}),B(x_{2},y_{2})  and large C(x_{3},y_{3}) as shown in figure below.

Draw the ordinates AL, CM and BN.

Note that area of triangle ABC = Area of trapezium ALNB + Area of trapezium ACML - Area of trapezium BCMN

= large frac{1}{2}(AL+BN)LN+frac{1}{2}(AL+CM)LM-frac{1}{2}(BN+CM)MN

But   large AL=y_{1},BN=y_{2},CM=y_{3}

  large LM=OM-OL=x_{3}-x_{1},MN=OM-ON=x_{3}-x_{2}

large and;LN=OL-ON=x_{1}-x_{2}

large therefore  Area of triangle ABC

  large =frac{1}{2}(y_{1}+y_{2})(x_{1}-x_{2})+frac{1}{2}(y_{1}+y_{3})(x_{3}-x_{1})-frac{1}{2}(y_{2}+y_{3})(x_{3}-x_{2})

large =frac{1}{2}[(y_{1}+y_{2})x_{1}-(y_{1}+y_{2})x_{2}+(y_{1}+y_{3})x_{3}-(y_{1}+y_{3})x_{1}-(y_{2}+y_{3})x_{3}+(y_{2}+y_{3})x_{2}]large =frac{1}{2}[(y_{1}+y_{2})x_{1}-(y_{1}+y_{3})x_{1}+(y_{2}+y_{3})x_{2}-(y_{1}+y_{2})x_{2}+(y_{1}+y_{3})x_{3}-(y_{2}+y_{3})x_{3}]

large =frac{1}{2}[x_1(y_{1}+y_{2}-y_{1}-y_{3})+x_2(y_{2}+y_{3}-y_{1}-y_{2})+x_3(y_{1}+y_{3}-y_{2}-y_{3})]

large =frac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]

 

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Sample Questions
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Question : 1

For what value of m the points ( -3, -5), (-5, -6) and (m, -4) are collinear ?

Right Option : C
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Explanation
Question : 2

Vertices of a triangle are (-2,0),(3,0) and (0,5). Area of this triangle is ______________ (in sq. units)

Right Option : B
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Explanation
Question : 3

For what value of 'n' the area of the triangle formd with points (1, n), (3, 5) and (-1, 2) will be 5 sq units ?

Right Option : A
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Explanation
 
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