Misc 8 - Find area of triangle formed by y - x = 0, x + y = 0

Misc 8 - Chapter 10 Class 11 Straight Lines - Part 2
Misc 8 - Chapter 10 Class 11 Straight Lines - Part 3
Misc 8 - Chapter 10 Class 11 Straight Lines - Part 4

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Let point where line (1) & (3) intersect be A & point where line (2) & (3) intersect be B We need to find Area triangle Δ OAB We know O(0,0) We need to find coordinates of A & B Coordinate of point A Point A is the intersection of x = k & y – x = 0 Putting x = k in equation (1) y − x = 0 y – k = 0 y = k So, point A (k, k) Coordinate of point B Point B is the intersection of x = k & y + x = 0 Putting x = k in equation (2) y + x = 0 y + k = 0 y = –k So, point B (k, –k) We know that Area of triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is = 1/2 |𝑥1(𝑦2 − 𝑦3) + 𝑥2(𝑦3 − 𝑦1) + 𝑥3 (𝑦1 − 𝑦2)| For ∆ AOB, (x1, y1) = O(0, 0) (x2,y2) = A(k, k) (x3,y3) = B(k, –k) Area of triangle ∆OAB whose vertices are (0, 0) (k, k) & (k, − k) = 1/2 |0(𝑘−(− 𝑘))+𝑘(−𝑘 −0) 𝑘(0−𝑘)| = 1/2 |0(k−(− k))+k(−k −0) k(0−k)| = 1/2 |0+k(−k)+k(−k)| = 1/2 |−k2−k2| = 1/2 |−2k2| = 2k2/2 = k2 square units Hence, required area of triangle is k2 square units

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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.