Question
If the points P(2, 3), Q(5, a) and R(6, 7) are
co-linear, then find the value of 'a'.Solution
If 3 given points are collinear, then the area of the triangle is zero. For P(x1, y1), Q(x2, y2) and R(x3, y3), the area of the triangle is So, 1/2 X [x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)] = 0 For the given set of points, area of the triangle = (1/2) X [2(a - 7) + 5(7 - 3) + 6(3 - a)] = 0 Or, (1/2) X [2a - 14 + 20 + 18 - 6a] = 0 Or, (1/2) X (24 - 4a) = 12 - 2a = 0 So, 2a = 12 And a = (12/2) = 6
I. 8y2 - 2y - 21 = 0
II. 2x2 + x - 6 = 0
I. 9/(4 )p + 7/8p = 21/12
II. 7/5p = 9/10q + 1/4
I. x² – 44x + 468 = 0
II. y² – 30y + 216 = 0
I. 96y² - 76y – 77 = 0
II. 6x² - 19x + 15 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 40x + 300 = 0
Equation 2: y² - 30y + 216 = 0
Find the roots of the equation 6p² – 5p – 6 = 0.
I. 10p² + 21p + 8 = 0
II. 5q² + 19q + 18 = 0
I. 2x² - 11x + 12 = 0
II. 12y² + 29y + 15 = 0
- What should be the value of t in the equation x² + tx + 64 = 0 so that it has two equal real roots?
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...