A(2, 7), B(4, -1), C(-2, 6) AB = √ (4 + 64) = √ 68 BC = √ (36 + 49) = √ 85 AC = √ (16+1) = √ 17 We see that AB² + AC² = BC² (follows Pythagoras theorem) ⇒ The triangle is a right triangle at A(2, 7).
I. 40 x² - 93 x + 54 = 0
II. 30 y² - 61 y + 30 = 0
1.3wx = 40 – wy
2. b2 = 2b + p
3. d2 + d = q
Now, observe the given conditions:
One root of equation...
I). 3p 2 – 13p + 12 = 0
II). 4q 2 + 3q – 7 = 0
(i) 2x² + 14x - 16 = 0
(ii) y² – y – 12 = 0
I. 64x2 - 64x + 15 = 0
II. 21y2 - 13y + 2 =0
I. x² + 4x + 4 = 0
II. y² - 8y + 16 = 0
I. (x13/5 ÷7) = 5488 ÷ x7/5
II. (y2/3 × y2/3 ) ÷ √4 = (343y)1/3...
I. 5x + 2y = 31
II. 3x + 7y = 36
I. 5x² + 17x + 6 = 0
II. 2y² + 11y + 12 = 0
...I. y² - 7 y – 18 = 0
II. x² + 10 x + 16 = 0