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JEE Maths Quiz on Coordinate Geometry
JEE Maths Quiz on Coordinate Geometry : In this article you will get to Online test for JEE Main, JEE Advanced, UPSEE, WBJEE and other engineering entrance examinations that will help the students in their preparation. These tests are free of cost and will useful in performance and inculcating knowledge. In this post we are providing you quiz on Coordinate Geometry
Quiz on Coordinate Geometry
Q1. The graph of the function cosx cos (x + 2) − cos^{2 }(x + 1) is
a) A straight line passing through (0, −sin^{2 }1) with slope 2
b) A straight line passing through (0, 0)
c) A parabola with vertex 75^{o}
d) A straight line passing through the point (π / 2, −sin^{2 }1) and parallel to the xaxis
d) A straight line passing through the point (π / 2, −sin^{2 }1) and parallel to the xaxis
Solution:
y = cos (x + 1 −1) cos (x + 1 + 1) − cos^{2 }(x + 1)
= cos^{2 }(x + 1) −sin^{2 }1 − cos^{2 }(x + 1) = −sin^{2 }1, which represents a straight line parallel to xaxis with y = −sin^{2 }1 for all x and so also for x = π / 2.
Q2. The area enclosed within the curve x+y= 1 is ____________.
a) 4
b) 2
c) 2
d) 6
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 √1^{2} + 1^{2} = √2
Hence, the area of square is √2 * √2 = 2 sq.units
♥ Trick: Required area = 2c^{2} / ab = [2 * 1^{2}] / [1 * 1] = 2.
Q3. The points (1, 3) and (5, 1) are the opposite vertices of a rectangle. The other two vertices lie on the line y=2x+c, then the value of c will be __________.
a) 4
b) 2
c) 2
d) 4
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.
Q4. The equations of two equal sides of an isosceles triangle are 7x − y + 3 = 0 and x + y − 3 = 0 and the third side passes through the point (1, 10). The equation of the third side is ___________.
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.
Q5. Aline through A (−5, − 4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0 and x − y − 5 = 0 at B, C and D, respectively. If (15 / AB)^{2 }+ (10 / AC)^{2} = (6 / AD)^{2}, then the equation of the line is _________.
a) 2x + 3y + 22 = 0
b) 3x + 2y + 22 = 0
c) 2x + 4y + 20 = 0
d) 4x + 5y + 20 = 0
a) 2x + 3y + 22 = 0
Soultion:
[x + 5] / [cosθ] = [y + 4] / [sinθ] = r_{1} / AB = r_{2} / AC = r_{3} / AD
(r_{1 }cosθ − 5, r_{1 }sinθ − 4) lies on x + 3y + 2 = 0
r_{1 }= 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
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