JEE Main Previous year Cartesian System and Straight Lines solutions offer greater insight into all the questions and the detailed answers provided here will further help students to take their preparation to a higher level. In this article, JEE aspirants can practice previous years’ questions on Cartesian System and Straight Lines asked in IIT JEE exams. The solutions have been prepared by our experts and they have provided accurate answers to all the listed questions which students can refer to and be ready to solve questions at the time of their examinations. We have included all the important maths questions and a detailed solution for each question. About 2-3 questions are asked from this chapter in the examination.
The important concepts such as distance between two points, coordinates of the point dividing any line segment, area of a triangle, the inclination of line, the angle between two lines, equation of the horizontal line, equation of the vertical line, a point-slope form of a line, the two-point form of a line, slope-intercept form of a line, the normal form of a line, the distance of a point from a line and perpendicular distance between two lines are covered here.
JEE Main Cartesian System and Straight Lines Past Year Questions With Solutions
Question 1: ABC is an isosceles triangle. If the coordinates of the base are B (1,3) and C (- 2, 7). The coordinates of vertex A can be.
(a) (1, 6)
(b) (1/2, 5)
(c) (5/6, 6)
(d) none of these
Solution:
The vertex A (x, y) is equidistant from B and C because ABC is an isosceles triangle.
Therefore, (x – 1)2 + (y – 3)2 = (x + 2)2 + (y – 7)2
⇒ 6x – 8y + 43 = 0
Check the options. Point (5/6, 6) satisfies the above equation.
Hence option c is the answer.
Question 2: If the vertices P, Q, and R of a triangle PQR are rational points, which of the following points of the triangle PQR is (are) always rational point(s)?
A) Centroid
B) Incentre
C) Circumcentre
D) Orthocentre (A rational point is a point both of whose coordinates are rational numbers)
Solution:
If A = (x1, y1), B = (x2, y2) and C = (x3, y3) where (x1, y1) are rational numbers then ∑x, ∑y are also rational.
So, the coordinates of the centroid will be rational.
As
Hence option A, C, D are correct.
Question 3: The new coordinates of a point (4, 5), when the origin is shifted to the point (1, -2) are
A) (5, 3)
B) (3, 5)
C) (3, 7)
D) None of these
Solution:
We know that if the origin is shifted to (h, k), then new coordinates (x, y) becomes (x – h, y – k). Therefore, the new coordinate of (4, 5) with respect to new origin (1, – 2) is (3, 7).
Hence option D is the answer.
Question 4: If P = (1, 0), Q = (-1, 0) and R = (2, 0) are three given points, then the locus of a point S satisfying the relation (SQ)2 + (SR)2 = 2(SP)2 is
A) A straight line parallel to the x-axis
B) A circle through the origin
C) A circle with centre at the origin
D) A straight line parallel to the y-axis
Solution:
Let S (x, y), then
⇒ 2x + 3 = 0
⇒ x = -3/2
Hence it is a straight line parallel to the y-axis.
Question 5: The coordinates of the points O, A and B are (0, 0), (0, 4) and (6, 0), respectively. If a points P moves such that the area of ΔPOA is always twice the area of ΔPOB, then find the equation to both parts of the locus of P.
Solution:
The three given points are O(0, 0), A(0, 4) and B(6, 0) and let P (x, y) be the moving point.
Area of ΔPOA = 2 * Area of ΔPOB
Hence, the equation to both parts of the locus of P is (x – 3y), (x + 3y) = 0.
Question 6: Find the Orthocentre of the triangle formed by the lines x + y = 1 and xy = 0.
Solution:
Given lines are x + y = 1 and xy = 0 when x = 0, then y = 1 when x = 1, then y = 0
∴ (0, 1) and (1, 0) are the vertices of triangle.
Clearly, the triangle is a right-angled isosceles.
Orthocentre of a right-angled triangle is the same as the vertex of the right angle.
Therefore, the point of intersection of x + y = 1 and xy = 0 is (0, 0).
Question 7: Find the circumcentre of a triangle formed by the line xy + 2x + 2y + 4 = 0 and x + y + 2 = 0.
Solution:
xy + 2x + 2y + 4 = 0 …..(i) and x + y + 2 = 0 ……(ii)
From (i), x (y + 2) + 2(y + 2) = 0
⇒ (x + 2) (y + 2) = 0
From (ii), x + y = -2
Vertices of triangle are (−2, 0) (-2, -2) (0, −2) (In a right angled triangle circumcentre is mid point of hypotenuse)
(−1, −1) is the circumcircle.
Question 8: Find the area of the triangle formed by the points (a,b + c), (b, c + a), (c, a + b).
Solution:
Area of triangle
Question 9: Three points are A (6, 3), B (−3, 5), C (4, −2) and P (x, y) is a point, then the ratio of the area of ΔPBC and ΔABC is
Solution:
Question 10: If P (1, 2), Q (4, 6) R (5, 7) and S (a, b) are the vertices of a parallelogram PQRS, then find the values of a and b.
Solution:
Diagonals cut each other at middle points. Hence,
Question 11: The locus of the point P (x, y) satisfying the relation
Solution:
Squaring both sides.
Again, squaring we get the equation of a pair of straight lines.
Question 12: The equations to a pair of opposite sides of a parallelogram are x2 – 5x + 6 = 0 and y2 – 6y + 5 = 0. Find the equations of its diagonals.
Solution:
Given x2 – 5x + 6 = 0
⇒ x = 2, 3
y2 – 6y + 5 = 0
⇒ y = 1, 5
Let ABCD be the parallelogram.
Vertices of the parallelogram are A(2, 1), B(3, 1), C(3, 5), D(2, 5).
Equation of diagonal d1 is
Equation of diagonal d2 is
So equations are, 4x + y = 13 and y = 4x − 7.
Question 13: The area bounded by the angle bisectors of the lines x2 – y2 + 2y = 1 and the line x + y = 3 is
Solution:
The angle bisectors of the lines x2 – y2 + 2y = 1 are x = 0 and y = 1.
Therefore, required area
Question 14: The lines represented by the equation ax2 + 2hxy + by2 + 2gx + 2gx + 2fy + c = 0 will be equidistant from the origin, if
Solution:
Let the equations represented by
On comparing, we get
According to the condition, the length of perpendiculars drawn from the origin to the lines are the same.
So,
Now on eliminating l, m, l′, m′and n, n′, we get the required condition
Also Read
JEE Advanced Maths Straight Lines Previous Year Questions with Solutions
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