Miscellaneous
Misc 2 Important
Misc 3
Misc 4
Misc 5 Important
Misc 6
Misc 7 Important You are here
Misc 8 Important
Misc 9
Misc 10 Important
Misc 11 Important
Misc 12
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Misc 19 Important
Misc 20 Important
Misc 21 Important
Misc 22
Misc 23 Important
Question 1 Important Deleted for CBSE Board 2025 Exams
Last updated at April 16, 2024 by Teachoo
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