Mathematics MCQ on Coordinate Geometry for JEE and Engineering Exam 2022
MCQ on Coordinate Geometry: To be an expert in JEE Mathematics, it is absolutely necessary to practice and be familiar will all the concepts as well as the questions of different types. This is essential to gain mastery over the subject. We have also often heard the common saying, “Practice Makes a Man Perfect”, hence students have to practice, practice and practice till they master the subject.
In this post we are providing you MCQ on Coordinate Geometry, which will be beneficial for you in upcoming JEE and Engineering Exams.
MCQ on Coordinate Geometry
Q1. In what direction can a line be drawn through the point (1, 2) so that its points of intersection with the line x + y = 4 is at a distance √6 / 3 from the given point?
a) θ = 35^{o} or 55^{o}
b) θ = 15^{o} or 75^{o}
c) θ = 25^{o} or 65^{o}
d) None of the above
Solution:
Let the required line through the point (1, 2) be inclined at an angle θ to the axis of x.
Then its equation is [x − 1] / [cosθ] = [y − 2] / [sinθ] = r …..(i)
where r is the distance of any point (x, y) on the line from the point (1, 2).
The coordinates of any point on the line (i) are (1 + r cosθ, 2 + r sinθ).
If this point is at a distance √6 / 3 form (1, 2), then r = √6 / 3.
Therefore, the point is (1 + [√6 / 3] cosθ, 2 + [√6 / 3] sinθ).
But this point lies on the line x + y = 4.
√6 / 3 (cosθ + sinθ) = 1 or
sinθ + cosθ = 3 / √6
[1 / √2] sinθ + [1 / √2] cosθ = √3 / 2, {Dividing both sides by √2}
sin (θ + 45^{o}) = sin60^{o} or sin 120^{o}
θ = 15^{o} or 75^{o}
Q2. If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is ________.
Solution:
Required locus of the point (x, y) is the curve|x| + |y| = 1.
If the point lies in the first quadrant, then x > 0, y > 0 and so |x| + |y| = 1
⇒ x + y = 1, which is straight line AB.
If the point (x, y) lies in the second quadrant then x < 0, y > 0 and so |x| + |y| = 1
⇒ −x + y = 1.
Similarly, for the third and fourth quadrant, the equations are −x −y = 1 and x − y = 1.
Hence, the required locus is the curve consisting of the sides of the square.
Q3. A variable line passes through a fixed point P. The algebraic sum of the perpendicular drawn from (2, 0), (0, 2) and (1, 1) on the line is zero, then what are the coordinates of the P?
a) P (1,1)
b) p(-1,1)
c) p (-1,-1)
d) none
Solution:
Let P (x_{1}, y_{1}), then the equation of the line passing through P and whose gradient is m, is y − y_{1} = m (x − x_{1}).
Now according to the condition
[{−2m + (mx_{1} − y_{1})} / {√1 + m^{2}}]+ [{2 + (mx_{1} − y_{1})} / {√1 + m^{2}} + {1 − m + (mx_{1} − y_{1})} / {√1 + m^{2}}=0
3 − 3m + 3mx_{1 }− 3y_{1} = 0
⇒ y_{1 }− 1 = m (x_{1} − 1)
Since it is a variable line, so hold for every value of m.
Therefore, y_{1 }= 1, x_{1 }= 1
⇒ P(1,1)
Q4. 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)-4
c) 1
d) none
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.
Q5. A-line 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) 2x + 3y + 22 = 0
c) 2x + 3y + 22 > 0
d) 2x + 3y – 22 < 0
[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
Q6. The line 2x + 3y = 12 meets the x-axis at A and y-axis at B. The line through (5, 5) perpendicular to AB meets the x-axis, y-axis and the AB at C, D and E, respectively. If O is the origin of coordinates, then the area of OCEB is _______.
a) [18 / 3] sq.units
b) [20 / 3] sq.units
c) [23 / 3] sq.units
d) [25 / 3] sq.units
Here O is the point (0, 0).
The line 2x + 3y = 12 meets the y-axis at B and so B is the point (0, 4).
The equation of any line perpendicular to the line 2x + 3y = 12 and passes through (5, 5) is 3x − 2y = 5 ……(i)
The line (i) meets the x-axis at C and so coordinates of C are (5 / 3, 0).
Similarly, the coordinates of E are (3, 2) by solving the line AB and (i).
Thus, O (0, 0), C (5 / 3, 0), E (3, 2) and B (0, 4).
Now the area of figure OCEB = area of ΔOCE + area of ΔOEB = [23 / 3] sq.units.
Q7. If the slope of a line passing through the point A (3, 2) be 3 / 4, then the points on the line which are 5 units away from A, are ________.
a) (1, 1) and (7, 3).
b) (1, 1) and (4, 5).
c) (1, -1) and (6, 5).
d) (1, 1) and (7, 5).
The equation of line passes through (3, 2) and of slope 3 / 4 is 3x − 4y − 1 = 0
Let the point be (h, k)then
3h − 4k − 1 = 0 …….(i) and
(h − 3)^{2} + (y − 2)^{2} = 5^{2} (ii)
On solving the equations, we get h = −1, 7 and k = −1, 5.
Hence, points are (1, 1) and (7, 5).
Q8. 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 ___________.
a) 3x + y + 7 = 0
b) x − 3y − 31 = 0.
c) Both a &b
d) None of the above
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.
Q9. The area enclosed within the curve |x|+|y|= 1 is ______.
a) 2
b)4
c) 5
d) 3
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.
Q10. The equation of the lines, which passes through the point (3, – 2) and are inclined at 60^{o} to the line √3x + y = 1 _____.
Solution:
The equation of any straight line passing through (3, 2) is y + 2 = m (x − 3) …….(i) The slope of the given line is −√3.
So, tan 60^{o }= ± [m − (−√3)] / [1 + m (−√3)]
On solving, we get m = 0 or √3
Putting the values of m in (i), the required equation of lines are
y + 2 = 0 and √3x − y = 2 + 3√3.
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