# Example 15 - Chapter 7 Class 10 Coordinate Geometry (Term 1)

Last updated at Aug. 16, 2021 by Teachoo

Examples

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Example 2 Important

Example 3 Important

Example 4

Example 5 Important

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Example 7 Important

Example 8

Example 9 Important

Example 10 Important

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Example 12 Deleted for CBSE Board 2022 Exams

Example 13 Important Deleted for CBSE Board 2022 Exams

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Example 15 Deleted for CBSE Board 2022 Exams You are here

Last updated at Aug. 16, 2021 by Teachoo

Example 15 If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD. Let the vertices of quadrilateral be A(−5, 7) , B(−4, −5) C(−1, −6) , D(4, 5) Joining AC There are 2 triangles formed ABC & ACD Hence, Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ADC Finding area ∆ ABC Area of triangle ABC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = −5 , y1 = 7 x2 = −4 , y2 = −5 x3 = −1 , y3 = −6 Putting values Area of triangle ABC = 1/2 [ −5(−5 –(−6)) + (−4)(−6 – 7) + (−1)(7 – (−5)) ] = 1/2 [ −5(−5 + 6) – 4(−13) + (−1)(7 + 5)] = 1/2 [ −5(1) – 4(−13) + (−1)(12)] = 1/2 [−5 + 52 − 12] = 𝟏/𝟐 [35] square units Similarly, Finding area ∆ ADC Area of triangle ADC = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = −5, y1 = 7 x2 = 4, y2 = 5 x3 = −1, y3 = −6 Area of triangle ADC = 1/2 [ −5(5 –(−6)) + 4(−6 – 7) + (−1)(7 – 5) ] = 1/2 [ −5(5 + 6) + 4(−13) + (−1)(2)] = 1/2 [ −5(11) + 4(−13) + (−1)(2)] = 1/2 [ −55 − 52 – 2] = 1/2 [ −109] But area cannot be negative, ∴ Area of triangle ADC = 𝟏/𝟐 [ 109] square units Area of quadrilateral ABCD = Area of triangle ABC + Area of triangle ADC = 1/2 [ 35 + 109] = 1/2 [ 144] = 72 square units