Area of Triangle when coordinates are given

Chapter 7 Class 10 Coordinate Geometry
Serial order wise

### Transcript

Ex 7.3 , 5 You have studied in Class IX, (Chapter 9, Example 3), that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A(4, – 6), B(3, –2) and C(5, 2) Let ∆ ABC be as shown in the figure Let AD be the median which divides BC into two equal parts, BD & CD Hence, Coordinates of D = ((𝑥1 + 𝑥2)/2 ", " (𝑦1 +𝑦2)/2) = ((3 + 5)/2 ", " (−2 + 2)/2) = (8/2 ", " 0/2) = (4, 0) Now, we need to prove that Area ∆ ABD = Area ∆ ACD Solving L.H.S Area of triangle ABD = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 4, y1 = −6 x2 = 3, y2 = −2 x3 = 4, y3 = 0 Putting values Area of triangle ABD = 1/2 [4(−2 − 0) + (3)(0 – (−6)) + 4(−6 – (−2))] = 1/2 [ 4(−2) + 3(0 + 6) + 4(−6 + 2)] = 1/2 [ 4(−2) + 3(6) + 4(−4)] = 1/2 [−8 + 18 −16] = 1/2 [−6] = −3 Now, since area cannot be negative Area of triangle ABC = 3 square units Solving R.H.S Area of triangle ACD = 1/2 [ x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2) ] Here x1 = 4 , y1 = −6 x2 = 5 , y2 = 2 x3 = 4 , y3 = 0 Area of triangle ACD = 1/2 [ 4(2 − 0) + (5)(0 – (−6)) + 4(−6 – 2) ] = 1/2 [ 4(2) + 5(0 + 6) + 4(−8) ] = 1/2 [ 4(2) + 5(6) + 4(−8) ] = 1/2 [ 8 + 30 − 32 ] = 1/2 [6] = 3 square units = R.H.S Hence, Area of ∆ ABD = Area of ∆ ACD Hence proved

#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.