# Misc 8

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 8 Find the area of the triangle formed by the lines y – x = 0, x + y = 0 and x – k = 0. There are three lines given in the graph y – x = 0 (0, 0) & (1, 1) satisfy this line x + y = 0 (0,0) & (1, –1) satisfy this line x – k = 0 (k, 0) & (k, 1) satisfy this line 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 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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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .