JEE Straight Line Previous Year Questions With Solutions

Straight line is the smallest distance between two points which is not curved in any direction and continuous in the same direction. Straight line is one of the important concepts in JEE syllabus. We at BYJU’S understand the importance of well-explained solution modules, hence, we have come up with an ultimate set of previous year question bank on topic straight lines. Straight lines solutions are available for students who are looking to solve questions from JEE past year papers. Students can get step-wise detailed explanation of all the problems in a very simple and easy manner on concepts, the perpendicular distance between two lines, point dividing a line segment, normal form of a line, area of a triangle, the point-slope form of a line, two-point form of a line, etc. About 2-3 questions are always asked from this chapter in the JEE exam.

Points to Remember:

JEE Main Past Year Questions With Solutions on Straight Lines

Questions given below have been asked in the past year’s JEE question papers.

Question 1: If C be the centroid of the triangle having vertices (3,−1), (1,3) and (2, 4). Let P be the point of intersection of the lines x + 3y − 1 = 0 and 3x − y + 1 = 0, then the line passing through the points C and P also pass through the point:

(a) (-9, -7)

(b) (-9, -6)

(c) (7, 6)

(d) (9, 7)

Answer: (b)

Solution:

Coordinates of C

Point of intersection of two lines: x + 3y – 1 = 0 and 3x – y + 1 = 0 is A(-1/5, 2/5)

slope, m = 8/11

Now, equation of line CA is y – 2 = (8/11)(x – 2)   (use equation y-y₁ = m(x-x₁))

8x – 11y + 6 = 0

and Point (-9, -6) lies on CP.

Question 2: The locus of mid points of the perpendiculars drawn from points on the line x = 2y to the line x = y is:

(a) 2x – 3y = 0

(b) 3x − 2y = 0

(c) 5x − 7y = 0

(d) 7x − 5y = 0

Answer: (c)

Solution:

Let R be the midpoint of PQ

PQ is perpendicular to line: y = x

Therefore, the equation of the line PQ can be written as

y = −x + c and it intersects y = x at point Q and x = 2y at point P.

Let R(h, k) be the midpoint of PQ.

(c, c) and (2a, a) be the Coordinates of points Q and P, respectively.

Now, Midpoint R = ((2a+c)/2, (a + c)/2)

slope of PQ = -1

(a-c)/(2a-c) = -1

c-2a = a – c

=> 2c = 3a

=> c = 3a/2

so h = 7a/4, k = 5a/4

h/k = 7/5

so 7y = 5x

This implies, 5x − 7y = 0

Question 3: Find the equation of line through (-4, 1, 3) and parallel to the plane x + y + z = 3 while the line intersects another line whose equation is x + y – z = 0 = x + 2y – 3z + 5

Answer: (c)

Solution:

The family of planes containing the line of intersection of planes is A + λB = 0

Where A = (x + y – z) and B = (x + 2y – 3z + 5)

That is,

(x + y – z) + λ(x + 2y – 3z + 5) = 0

The above line passes through the point (-4, 1, 3), so satisfy the equation.

On substituting the above equation, we get

λ = -1

=> Equation of plane is y – 2z + 5 = 0

The required line lies in this plane and is parallel to x + y + z = 5.

Now,

Question 4: A plane parallel to y-axis passing through line of intersection of planes x + y + z = 1 and 2x + 3y – z – 4 = 0 which of the point lie on the plane.

(a) (3, 2, 1)

(b) (-3, 0, 1)

(c) (-3, 1, 1)

(d) (3, -1, 1)

Answer: (d)

Solution:

Required plane is A + λB = 0

Where A = (x + y + z – 1) and B = (2x + 3y – z – 4)

That is,

(x + y + z – 1) + λ(2x + 3y – z – 4) = 0

=> (1 + 2λ)x + (1 + 3λ)y + (1 – λ)z – (1 + 4λ) = 0

As the above equation is parallel to the y-axis, so λ = -1/3

Therefore, required plane is x + 4z + 1 = 0

Substitute and check all options,

we get (3, -1, 1) is the point which lie in this plane.

Question 5: If the foot of the perpendicular drawn from the point (1, 0, 3) on a line passing through (α, 7, 1) is (5/3, 7/3, 17/3), then What will be the value of α.

Solution:

Let P(1, 0, 3) and Q(5/3, 7/3, 17/3) are given points.

Direction ratios of line L is:

(α – 5/3, 7 – 7/3, 1 – 17/3) = ( (3α – 5)/3, 14/3, -14/3 )

Direction ratios of PQ: (-2/3, -7/3, -8/3)

Since line L is perpendicular to PQ then

=> -6α +10 – 98 + 112 = 0

=> 6α = 24

=> α = 4

Therefore, the value of α is 4.

Question 6: If the line, (x-3)/1 = (y+2)/-1 = (z+λ)/-2 lie in the plane, 2x – 4y + 3z = 2, then the shortest distance between this line and the line (x-1)/12 = y/9 = z/4 is:

(a) 2

(b) 1

(c) 0

(d) 3

Answer: (c)

Solution:

Given line (x-3)/1 = (y+2)/-1 = (z+λ)/-2 = k

Let P be any point on the plane,

P = (k + 3, -k – 2, -2k – λ )

P lie on the plane, 2x – 4y + 3z = 2, Which means P satisfy the equation of the plane.

So, 2(k + 3) – 4(-k – 2) + 3(-2k – λ) = 2

Solving above equation, we have λ = 4

Now, the shortest distance between the below lines:

(x-3)/1 = (y+2)/-1 = (z+4)/-2 and (x-1)/12 = y/9 = z/4 is

= |2(14) + 2(28) – 4(21)|

= 0

=> d² = 0

or d = 0

Question 7: Let A(1, 0), B(6, 2), C(3/2, 6) be the vertices of a triangle ABC. If P is a point inside the triangle ABC such that the triangles APC, APB and BPC have equal areas, then find the length of the line segment PQ, where 𝑄 is the point (-7/6, -1/3).

Solution:

Since point P is the centroid, so its coordinates are

and coordinates of point Q : (-7/6, -1/3)

Now, PQ = √[4²+3²] = 5   (using distance formula)

Question 8: Let two points be A(1, −1) and B(0, 2). If a point P(x’, y’) be such that the area of triangle PAB = 5 sq. units and it lies on the line, 3x + y – 4λ = 0, then the value of λ is

(a) 4

(b) 1

(c) -3

(d) 3

Answer: (d)

Solution:

Let two points be A(1, −1) and B(0, 2). If a point P(x’, y’) be such that the area of triangle PAB = 5 sq. units

Area of triangle is:

=> (2 – y’) – x’ – 2x’ = ± 10

=> 3x’ + y’ = 12 or 3x’ + y’ = -8

Since triangle lie on the line 3x + y – 4λ = 0

comparing we get

=> λ = 3 or λ = -2

Neglecting negative value.

So, λ = 3

Hence option d is the answer.

Question 9: Find the shortest distance between the lines:

Solution:

Question 10: The length of the perpendicular from the origin, on the normal to the curve, x² + 2xy – 3y² = 0 at the point (2,2) is:

(a) 2

(b) 2√2

(c) 4√2

(d) √2

Answer: (b)

Solution:

Given curve: x² + 2xy – 3y² = 0

=> x² + 3xy – xy – 3y² = 0

=> x(x + 3y) – y(x + 3y) = 0

=> (x + 3y)(x – y) = 0

=> x + 3y = 0 or x – y = 0

Since point (2, 2) lies on x – y = 0

Therefore,

Equation of normal will be x + y = λ

Put x = 2 and y = 2

So, λ = 4

Required line is x + y = 4

Now, Distance of L from the origin = |−4/√2 | = 2√2

Also Read

Straight Line JEE Advanced Previous Year Questions With Solutions

Straight Lines – Important JEE Main Questions