Misc 14 - Find area bounded by lines: 2x + y = 4, 3x-2y=6 - Miscellaneous

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Misc 14 Using the method of integration find the area of the region bounded by lines: 2 + = 4, 3 2 =6 and 3 +5=0 Step 1: Make Figure 2 + = 4 3 2 = 6 3 + 5 = 0 Step 2: Find intersecting Points A & B Point A Point A is intersection of lines x 3y + 5 = 0 & 2x + y = 4 Now, x 3y + 5 = 0 x = 3y 5 Putting x = 3y 5 in 2x + y = 4 2(3y 5) + y = 4 6y 10 + y = 4 7y = 14 y = 2 Putting y = 2 in x 3y + 5 = 0 x 3(2) + 5 = 0 x 6 + 5 = 0 x = 1 So, point A (1, 2) Point B Point B is intersection of lines x 3y + 5 = 0 & 3x 2y = 6 Now, x 3y + 5 = 0 x = 3y 5 Putting x = 3y 5 in 3x 2y = 6 3(3y 5) 2y = 6 9y 15 2y = 6 7y = 21 y = 3 Putting y = 3 in x 3y + 5 = 0 x 3(3) + 5 = 0 x 9 + 5 = 0 x = 4 So, point A (4, 3) Finding area Area ABED Area ABED = 1 4 Equation of AB 3 +5=0 +5=3 + 5 3 = = + 5 3 Therefore, Area ABED = 1 4 +5 3 = 1 3 1 4 +5 = 1 3 2 2 +5 1 4 = 1 3 4 2 2 +5.4 1 2 2 +5.1 = 1 3 8+20 1 2 5 = 1 3 45 2 = 15 2 Area ACD Area ACD = 1 2 Equation of line AC 2 + =4 =4 2 Area ACD = 1 2 4 2 = 4 2 2 2 1 2 = 4 2 1 2 = 4.2 2 2 4.1 1 2 = 8 4 4+1 = 1 Area CBE Area CBE = 2 4 Equation of line BC 3 +2 =6 3 6=2 3 6 2 = = 3 6 2 Therefore, Area CBE = 2 4 3 6 2 = 1 2 2 4 3 6 = 1 2 3 2 2 6 2 4 = 1 2 3.4 2 2 6.4 3.2 2 2 6.2 = 1 2 24 24 6+12 =3 Hence Area Required = Area ABED Area ACD Area CBE = 15 2 1 3 = 15 2 4 = 15 8 2 = 7 2 square units

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.