JEE Main Coordinate Geometry Important Questions

JEE Main Coordinate Geometry Important Questions: JEE Main 2024 Coordinate Geometry is one of the key chapters from the Mathematics section from which we can anticipate at least four questions in the upcoming JEE Main exam . The study of geometric structures and connections using coordinates allows for exact mathematical analysis of points, lines, curves, and conic sections in a plane. Solving coordinate geometry questions in the JEE Main test can be difficult because it requires good knowledge of geometric principles, algebraic approaches and a lot of practice. Therefore, we have compiled some of the JEE Main 2024 important questions with solutions for Coordinate Geometry in this post. Solving these JEE Main Coordinate Geometry questions will allow you to gain expertise in solving tough questions and develop problem-solving approaches.

NTA has announced the JEE Main exam 2024 dates. As per the schedule, the JEE Main Phase 1 exam will be held from January 24 to February 1, 2024, and Phase 2 will be conducted from April 1 to 15, 2024.

LATEST: JEE Main 2024 Daily Practice Questions for 25 September 2023

Read the full post to get access to JEE Main Coordinate Geometry important questions with solutions, JEE Main Coordinate Geometry previous year question papers PDF, and other important practice material.

JEE Main Coordinate Geometry Important Questions

Candidates can go through the JEE Main 2023 Coordinate Geometry important questions given below. Attempting these questions will help you gain expertise in solving tough questions from the mathematics subject and score good marks in the exam.

Q1: The graph of the function cos x cos (x + 2) − cos2 (x + 1) is

A) A straight line passing through (0, − sin2 1) with slope 2

B) A straight line passing through (0, 0)

C) A parabola with vertex 75°

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

Solution:

Let y = cos x cos (x + 2) − cos2 (x + 1)

= 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.

Q2: 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.

Q3: 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?

Solution:

Let the required line through the point (1, 2) be inclined at an angle θ to the x-axis.

Then its equation is [x − 1] / [cos θ] = [y − 2] / [sin θ] = 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 cos θ, 2 + r sin θ).

If this point is at a distance √6 / 3 form (1, 2), then r = √6 / 3.

Therefore, the point is (1 + [√6 / 3] cos θ, 2 + [√6 / 3] sin θ).

But this point lies on the line x + y = 4.

√6 / 3 (cos θ + sin θ) = 1 or

sin θ + cos θ = 3 / √6

[1 / √2] sin θ + [1 / √2] cos θ = √3 / 2, {Dividing both sides by √2}

sin (θ + 45o) = sin60o or sin 120o

θ = 15o or 75o

Q4:  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 _________.

Solution:

[x + 5] / [cos θ] = [y + 4] / [sin θ] = r1 / AB = r2 / AC = r3 / AD

(r1 cos θ − 5, r1 sin θ − 4) lies on x + 3y + 2 = 0

r1 = 15 / [cos θ + 3 sin θ]

Similarly, 10 / AC = 2 cos θ + sin θ and 6 / AD = cos θ − sin θ

Putting in the given relation, we get (2 cos θ + 3 sin θ)2 = 0

tan θ = −2 / 3

⇒ y + 4 = [−2 / 3] (x + 5)

2x + 3y + 22 = 0

Q 5: 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?

Solution:

Let P (x1, y1), then the equation of the line passing through P and whose gradient is m, is y − y1 = m (x − x1).

Now according to the condition

[{−2m + (mx1 − y1)} / {√1 + m2}]+ [{2 + (mx1 − y1)} / {√1 + m2}] + [{1 − m + (mx1 − y1)} / {√1 + m2}] = 0

3 − 3m + 3mx1 − 3y1 = 0

⇒ y1 − 1 = m (x1 − 1)

Since it is a variable line, hold for every value of m.

Therefore, y1 = 1, x1 = 1

⇒ P(1, 1)

Also Read: JEE Main 2024: Strategy to Score 90+ in Mathematics

Q 6: The locus of a point P, which divides the line joining (1, 0) and (2 cos θ, 2 sin θ) internally in the ratio 2 : 3 for all θ, is a ________.

Solution:

Let the coordinates of the point P, which divides the line joining (1, 0) and (2 cos θ, 2 sin θ) in the ratio 2 : 3 be (h, k). Then,

h = [4 cos θ + 3] / [5] and k = [4 sin θ] / [5]

cos θ = [5h − 3] / [4] and sin θ = [5k] / [4]

([5h − 3] / [4])2 + ([5k] / [4])2 = 1 {since cos2θ + sin2θ = 1}

(5h − 3)2 + (5k2) = 16

Therefore, locus of (h, k) is (5x − 3)2 + (5y)2 = 16, which is a circle.

Question 7: The area of a parallelogram formed by the lines ax ± by ± c = 0, is __________.

Solution:

ax ± by ± c = 0

⇒ x / [± c / a] + y/ [± c / b] = 1 which meets on axes at A (ca, 0), C (−ca, 0), B (0, cb), D (0, −cb).

Therefore, the diagonals AC and BD of quadrilateral ABCD are perpendicular.

Hence, it is a rhombus.

So, the area = (1/2) × AC × BD

= (1/2) × (2c/a) × (2c/b)

= 2c2 / ab.

Question 8: The area enclosed within the curve |x|+|y|= 1 is ____________.

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.

The length of each side of this square is √12 + 12 = √2

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

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

Also Read: JEE Main 2024 Maths Chapter-wise Topics with Weightage

JEE Coordinate Geometry Previous Year Paper PDF

Candidates can access the JEE Main Coordinate Geometry previous year question paper PDF from the link given below.

Benefits of Solving JEE Main Coordinate Important Questions

Solving JEE Main 2024 Coordinate Geometry important questions is beneficial for candidat’es preparation because

• These questions help you comprehend coordinate geometry ideas properly, which is necessary for doing well in this section of JEE Main exam.
• Practicing a range of JEE Main 2024 Coordinating important questions sharpens your problem-solving abilities, allowing you to efficiently solve complicated geometrical issues.
• Solving complicated questions of the JEE Main Mathematics section within a time restriction helps to enhance time management abilities, which are important in the JEE Main exam
• Regular preparation helps you recognize trends and popular question types, which improves your ability to strategize throughout the test
• Tracking your exam preparation by solving the JEE Main 2024 important questions helps you to discover weak points and focus on improvement.

JEE Main Mathematics Question Papers

Solving the JEE Main previous question paper will help the students know what kind of questions are asked in the Mathematics section, how to solve it, evaluate the performance and work on the mistakes.  Click on the links given below to access the JEE Main Mathematics question papers PDF

JEE Main Previous Year Question Papers

JEE Main exam in the previous year was conducted in two sessions. Candidates can click on the links given below and attempt the JEE Main previous year paper.

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