Maths Quiz on Coordinate Geometry for IIT JEE & Engineering Exam

d) A straight line passing through the point (π / 2, −sin² 1) and parallel to the x-axis

Solution:

y = cos (x + 1 −1) cos (x + 1 + 1) − cos² (x + 1)

= cos² (x + 1) −sin² 1 − cos² (x + 1) = −sin² 1, which represents a straight line parallel to x-axis with y = −sin² 1 for all x and so also for x = π / 2.

c) 2

Solution:

The given lines are ± x ± y = 1 i.e. x + y = 1, x − y = 1, x + y = −1 and x − y = −1.

These lines form a quadrilateral whose vertices are A (−1, 0), B (0, −1), C (1, 0) and D (0, 1) Obviously ABCD is a square.

Length of each side of this square is √1² + 1² = √2

Hence, the area of square is √2 * √2 = 2 sq.units

♥ Trick: Required area = 2c² / |ab| = [2 * 1²] / [|1 * 1|] = 2.

d) -4

Solution:

Let ABCD be a rectangle. Given A (1, 3) and C (5, 1).

Equation B and D lie on y=2x+c We know that the intersecting point of diagonal of a rectangle is the same or at the midpoint.

So, midpoint of AC is (3, 2).

Hence, y = 2x + c passes through (3, 2).

Therefore, c = −4.

Solution:

Any line through (1, 10) is given by y + 10 = m (x − 1)

Since it makes equal angle say α with the given lines 7x − y + 3 = 0 and x + y − 3 = 0, therefore tan α = [m − 7] / [1 + 7m]

= [m − (−1)] / [1 + m (−1)]

⇒ m = [1 / 3] or 3

Hence, the two possible equations of the third side are 3x + y + 7 = 0 and x − 3y − 31 = 0.

a) 2x + 3y + 22 = 0

Soultion:

[x + 5] / [cosθ] = [y + 4] / [sinθ] = r₁ / AB = r₂ / AC = r₃ / AD

(r₁ cosθ − 5, r₁ sinθ − 4) lies on x + 3y + 2 = 0

r₁ = 15 / [cosθ + 3 sinθ]

Similarly, 10 / AC = 2 cosθ + sinθ and 6 / AD = cosθ − sinθ

Putting in the given relation, we get (2 cosθ + 3 sinθ)² = 0

tan θ = −2 / 3

⇒ y + 4 = [−2 / 3] (x + 5)

2x + 3y + 22 = 0