JEE Main Previous Year Questions (2016- 2024): Straight Lines and Pair of Straight Lines PDF Download

Q.1. The equations of the sides ⁢B, BC and C⁢ of a triangle ⁢ABC are : 2x + y = 0, x + py = 21a, (a ≠ 0) and x − y = 3 respectively. Let P(2, a) be the centroid of Δ ⁢BC. Then (BC)² is equal to         (JEE Main 2023)

Ans.122


–2α + 2 – α = 3a + 2
α = –a
put 'B' in BC
α – 2pα = 21a
α.(1 - 2p) = 21a
2p - 1 = 21
p = 11
put 'C' in BC
β + 3 + 11β = 21a
21a + 12β +3 = 0
also β = 2 – α Solving a = –3, β = 5

BC² = 122

Q.2. Let A(α,−2),B(α, 6) and C(α/4, −2) be vertices of a ΔABC. If (5, α/4) is the circumcentre of ΔABC, then which of the following is NOT correct about ΔABC ?        (JEE Main 2022)
(a) area is 24
(b) perimeter is 25
(c) circumradius is 5
(d) inradius is 2

Ans.b

Circumcentre of ΔABC



Circumradius = 10/2 = 5
r = Δ/s = 24/12 = 2

Q.3. Let m₁, m₂ be the slopes of two adjacent sides of a square of side a such that a² + 11a + 3(m₁² + m₂²) = 220. If one vertex of the square is (10(cos⁡α − sin⁡α), 10(sin⁡α + cos⁡ α)), where α ∈(0, π/2) and the equation of one diagonal is (cos⁡α − sin⁡α)x + (sin⁡α + cos⁡α)y = 10, then 72(sin⁴⁡α + cos⁴⁡α) + a² − 3a + 13 is equal to:    (JEE Main 2022)
(a) 119
(b) 128
(c) 145
(d) 155

Ans.b

The slopes $m\_1$m1 and $m\_2$m2 of the adjacent sides satisfy:$m\_1\; \backslash cdot\; m\_2\; =\; -1,$m1⋅m2=−1,

because the sides are perpendicular.

The equation of the diagonal simplifies directly:

This matches the point:

Thus, the point satisfies the diagonal equation, confirming that the vertex lies on the diagonal.

Now, Analyze

Substitute values into:

After computation, the value is 128.$\backslash boxed\{128\}.$

Q.4. Let the circumcentre of a triangle with vertices A(a, 3), B(b, 5) and C(a, b), ab > 0 be P(1,1). If the line AP intersects the line BC at the point Q(k₁, k₂), then k₁ + k₂ is equal to:    (JEE Main 2022)
(a) 2
(b) 4/7
(c) 2/7
(d) 4

Ans.b

Let D be mid-point of AC, then

Let E be mid-point of BC,

On putting b=−1, we get a = 5 or −3
But a = 5 is rejected as ab > 0
A(−3,3), B(−1,5),C(−3, −1),P(1, 1)
Line BC ⇒ y = 3x + 8
Line AP ⇒ y = 3 − x/2
Point of intersection

Q.5. The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 39 and x − y = 3 respectively and P(2, 3) is its circumcentre. Then which of the following is NOT true?    (JEE Main 2022)
(a) (AC)² = 9p
(b) (AC)² + p² = 136
(c) 32 < area(ΔABC) < 36
(d) 34 < area(ΔABC) < 38

Ans.d
Intersection of 2x + y = 0 and x - y  = 3 : A(1, - 2)

Equation of perpendicular bisector of AB is
x − 2y = −4
Equation of perpendicular bisector of AC is
x + y = 5
Point B is the image of A in line x − 2y + 4 = 0 which is obtained as B(−13/5, 26/5)
Similarly vertex C : (7,4)
Equation of line BC : x + 8y = 39
So, p = 8
AC = (7−1)2+(4+2)2=62

Area of triangle ABC = 32.4

Q.6. Let A(1,1),B(−4,3),C(−2,−5) be vertices of a triangle ABC, P be a point on side BC, and Δ₁ and Δ₂ be the areas of triangles APB and ABC, respectively. If Δ₁ : Δ₂ = 4:7, then the area enclosed by the lines AP,AC and the x-axis is (JEE Main 2022)
(a) 1/4
(b) 3/4
(c) 1/2
(d) 1

Ans.c

Step 1: Total Area of △ABC:

We use the area formula for a triangle with vertices $(x\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">1</sub>},y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">1</sub>}),(x\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sub>},y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sub>}),(x\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sub>},y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sub>})(x\_1,\; y\_1),\; (x\_2,\; y\_2),\; (x\_3,\; y\_3)$:

$Area\; of\mathrm{Area\; of\; △}ABC=\frac{1}{2}\mid x\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">1</sub>}(y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sub>}-y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sub>})+x\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sub>}(y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sub>}-y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">1</sub>})+x\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">3</sub>}(y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">1</sub>}-y\mathrm{<sub\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">2</sub>})\mid .\backslash text\{Area\; of\; \}\; \backslash triangle\; ABC\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; x\_1(y\_2\; -\; y\_3)\; +\; x\_2(y\_3\; -\; y\_1)\; +\; x\_3(y\_1\; -\; y\_2)\; \backslash right|$

Substitute $A(1,1),\; B(-4,3),\; C(-2,-5)$A(1,1),B(−4,3),C(−2,−5):

$Area\; of$Area of △ABC=1 / 2 ∣1(3−(−5))+(−4)((−5)−1)+(−2)(1−3)∣.$Area\; of$

Area of △ABC= 1 / 2 ∣1(8)+(−4)(−6)+(−2)(−2)∣.

Step 2: Area of△APB:

The ratio of areas $\backslash Delta\_1\; :\; \backslash Delta\_2\; =\; 4\; :\; 7$Δ₁: Δ₂=4:7, so:

Step 3: Point P on Line Segment BC:

Let $\mathrm{<span\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;">P\; divide </span>}$$BC$BC in the ratio $k\; :\; 1$k:1. The coordinates of $P$P can be found using the section formula:

Step 4: Equation of Line AP:

The equation of a line passing through two points (x₁,y₁) and (x₂,y₂) is:

Step 5: Equation of Line AC:

For A(1,1) and $C(-2,-5)$C(−2,−5), the slope is:

The equation of $AC$AC is:

y−1=2(x−1).

$y\; =\; 2x\; -\; 2\; +\; 1\; =\; 2x\; -\; 1.$y=2x−2+1=2x−1.

Step 6: Find Intersection with the x-Axis:

• For $AP$AP, set $y\; =\; 0$y=0:

  $0=\frac{2}{3}x+\frac{1}{3}.$
  $\mathrm{<span\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;"\; fr-original-class="base"><span\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;"\; fr-original-class="mord\; mathnormal">x </span><span\; style="word-break:\; break-word;\; line-height:\; 1.8;\; font-family:\; Lato,\; sans-serif;\; font-size:\; 20px;\; letter-spacing:\; inherit;\; color:\; black;"\; fr-original-class="mrel">=\; 1/2</span></span>}$

  Intersection: $(-\frac{1}{2},0).$

  
  

Step 7: Area Enclosed by $AP,\; AC$AP,AC, and x-Axis:

The area is a triangle with vertices:

• $A(1,1)$A(1,1),
• Intersection of $AP$AP and x-axis: $(-\frac{1}{2},0)\backslash left(-\backslash frac\{1\}\{2\},\; 0\backslash right)$,
• Intersection of $AC$AC and $x$x-axis: $(\frac{1}{2},0)\backslash left(\backslash frac\{1\}\{2\},\; 0\backslash right)$.

$Area\; of$Q.7. A point P moves so that the sum of squares of its distances from the points (1,2) and (−2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A, B and the y-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to:        (JEE Main 2022)
(a)9/2
(b)
(c)
(d) 9

Ans.b
Let point P:(h,k)
(h − 1)² + (k − 2)² + (h + 2)² + (k − 1)² = 14
2h² + 2k² + 2h − 6k − 4 = 0
Locus of P : x² + y² + x − 3y − 2 = 0
Intersection with x-axis,
x² + x − 2 = 0
⇒ x = −2, 1
Intersection with y-axis,
y² − 3y − 2 = 0

Area of the quadrilateral

Q.8. Let the point P(α, β) be at a unit distance from each of the two lines L₁ : 3x − 4y + 12 = 0, and L₂ : 8x + 6y + 11 = 0. If P lies below L₁ and above L₂, then 100(α + β) is equal to        (JEE Main 2022)
(a) −14
(b) 42
(c) −22
(d) 14

Ans.d

L₁ : 3x − 4y + 12 = 0
L₂ : 8x + 6y + 11 = 0
Equation of angle bisector of L₁ and L₂ of angle containing origin
2(3x − 4y + 12) = 8x + 6y + 11
2x + 14y − 13 = 0 ...... (i)

⇒ 3α − 4β + 7 = 0 ...... (ii)
Solution of 2x + 14y − 13 = 0 and 3x − 4y + 7 = 0 gives the required point P(α,β),α = −23/25, β = 53/50
100(α + β) = 14

Q.9. A line, with the slope greater than one, passes through the point A(4, 3) and intersects the line x − y − 2 = 0  at the point B. If the length of the line segment AB is , then B also lies on the line : (JEE Main 2022)
(a) 2x + y = 9
(b) 3x − 2y = 7
(c) x + 2y = 6
(d) 2x − 3y = 3

Ans.c

Q.10. Let α₁, α₂ (α₁ < α₂) be the values of α fo the points (α, −3), (2, 0) and (1, α) to be collinear. Then the equation of the line, passing through (α₁, α₂) and making an angle of π/3 with the positive direction of the x-axis, is : (JEE Main 2022)
(a) x − √3y − 3√3 + 1 = 0
(b) √3x − y + √3 + 3 = 0
(c) x − √3y + 3√3 + 1 = 0
(d) √3x − y + √3 − 3 = 0

Ans.b

Points A(α, −3), B(2, 0) and C(1, α) are collinear.

∴ Slope of AB = Slope of BC

⇒ −3 = α(2 − α)
⇒ −3 = 2α − α²
⇒ α² − 2α − 3 = 0
⇒ α² − 3α + α − 3 = 0
⇒ α(α − 3) + 1(α − 3) = 0
⇒ (α + 1)(α − 3) = 0
⇒ α = −1, 3
Given, α₁< α₂
∴ α₁ = 1 and α₂ = 3
∴ (α₁, α₂) = (−1, 3)
Now, equation of the line passing through (−1, 3) and making angle π/3 with positive x-axis is
(y − y₁) = m(x − x₁)

Q.11. Let the area of the triangle with vertices A(1, α), B(α, 0) and C(0, α) be 4 sq. units. If the points (α, −α), (−α, α) and (α², β) are collinear, then β is equal to (JEE Main 2022)
(a) 64
(b) −8
(c) −64
(d) 512

Ans.c

∵ A(1, α), B(α, 0) and C(0, α) are the vertices of ΔABC and area of ΔABC = 4

⇒ |1(1 − α) − α(α) + α²| = 8
⇒ α = ±8
Now, (α,−α),(−α,α) and (α², β) are collinear

⇒ 8(8 − β) + 8(−8 − 64) + 1(−8β − 8 × 64) = 0
⇒ 8 − β − 72 − β − 64 = 0
⇒ β = - 64

Q.12. Let R be the point (3, 7) and let P and Q be two points on the line x + y = 5 such that PQR is an equilateral triangle. Then the area of ΔPQR is:        (JEE Main 2022)
(a)
(b)
(c)
(d)

Ans.d

Let, side of triangle = a.


From figure, h = a sin ⁡60^∘  

Q.13. If two straight lines whose direction cosines are given by the relations l + m − n = 0, 3l² + m² + cnl = 0 are parallel, then the positive value of c is : (JEE Main 2022)
(a) 6
(b) 4
(c) 3
(d) 2

Ans.a
l + m − n = 0 ⇒ n = l + m
3l² + m² + cnl = 0
3l² + m² + cl(l + m) = 0
= (3 + c)l² + clm + m² = 0

∴ Lines are parallel
D = 0
c² − 4(3 + c) = 0
c² − 4c − 12 = 0
(c − 4)(c + 3) = 0
c = 4 (as c > 0)

Q.14. In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If (α, β) is the centroid of ΔABC, then 15(α + β) is equal to:        (JEE Main 2022)
(a) 39
(b) 41
(c) 51
(d) 63

Ans.c


B(1, 2)
Let C(k, 4 − 2k)
Now AB² = AC²
5² + (−1)² = (6 − k)² + (−3 + 2k)²
⇒ 5k² −2 4k + 19 = 0
(5k − 19)(k − 1) = 0 ⇒ k = 19/5

Centroid (α, β)

Now 15(α + β)
15(17/5) = 51

Q.15. Let a triangle be bounded by the lines L₁ : 2x + 5y = 10; L₂ : −4x + 3y = 12 and the line L₃, which passes through the point P(2, 3), intersects L₂ at A and L₁ at B. If the point P divides the line-segment AB, internally in the ratio 1 : 3, then the area of the triangle is equal to : (JEE Main 2022)
(a) 110/13
(b) 132/13
(c) 142/13
(d) 151/13

Ans. b

Step 1: Find the intersection point C of lines $L\_1$L₁ and $L\_2$L₂

Equations of the lines:

Multiply through by 5 to eliminate the fraction:

Thus, the intersection point is: 

Step 2: Find the equation of line L₃

Line $L\_3$L₃ passes through P(2,3) and intersects $L\_1$L₁ at $B$B and $L\_2$L₂ at $A$A. Let $L\_3$L₃ have slope $m$m, so its equation is:

y−3=m(x−2) ⟹ y=mx−2m+3

Step 3: Find the intersection points A and $B$B

Intersection of $L\_3$L₃ with $L\_2$L₂ (Point $A$A):

Intersection of $L\_3$L₃ with $L\_1$L₁ (Point $B$B):

Step 4: Use the section formula to find coordinates of A and $B$B

Since $P(2,3)$P(2,3) divides $AB$AB in the ratio 1:3, the coordinates of $A$A and $B$B can be computed using the section formula.

Step 5: Calculate the area of $\backslash triangle\; ABC$△ABC

Using the formula for the area of a triangle given vertices A(x₁,y₁), B(x₂,y₂), and C(x₃,y₃):

$Area\backslash text\{Area\}\; =\; \backslash frac\{1\}\{2\}\; \backslash left|\; x\_1(y\_2\; -\; y\_3)\; +\; x\_2(y\_3\; -\; y\_1)\; +\; x\_3(y\_1\; -\; y\_2)\; \backslash right|.$Area = 1/ 2 ∣x₁(y₂−y₃)+x₂(y₃−y₁)+x₃(y₁−y₂)∣.

Substitute the coordinates of $A$A, B, and C to compute the area.

Q.16. The distance of the origin from the centroid of the triangle whose two sides have the equations x − 2y + 1 = 0 and 2x − y − 1 = 0 and whose orthocenter is (7/3, 7/3) is :        (JEE Main 2022)
(a) 2
(b) 2
(c) 22
(d) 4

Ans. c

For point A,
2x − y − 1 = 0 ...... (1)
x − 2y + 1 = 0 ...... (2)
Solving (1) and (2), we get
x = 1, y = 1.
∴ Point A = (1, 1)
Altitude from B to line AC is perpendicular to line AC.
∴ Equator of altitude BH is
2x + y + λ = 0 ...... (3)
It passes through point H(7/3, 7/3) so it satisfy the equation (3).

⇒ α = −7
∴ Altitude BH = 2x + y − 7 = 0 ...... (4)

Solving equation (1) and (4), we get
x = 2, y = 3.
∴ Point B = (2, 3)
Altitude from C to line AB is perpendicular to line AB.
∴ Equation of altitude CH is
x + 2y + λ = 0 ...... (5)
It passes through point H(7/3, 7/3) so it satisfy equation (5).

⇒ λ = −7
∴ Altitude CH = x + 2y − 7 = 0 ...... (6)
Solving equation (2) and (6), we get
x = 3, y = 2
∴ Point C = (3, 2)
Centroid G (x, y) of triangle A (1, 1), B (2, 3) and C (3, 2) is

Now d Distance of point G (2, 2) from center O (0, 0) is

Q.17. The distance between the two points A and A' which lie on y = 2 such that both the line segments AB and A' B (where B is the point (2, 3)) subtend angle π/4 at the origin, is equal to:        (JEE Main 2022)
(a) 10
(b) 48/5
(c) 52/5
(d) 3

Ans. c
Step 1: Identify the coordinates of points A and A′. 
Let the coordinates of point A be (x₁, 2)and the coordinates of point A′ be (x₂, 2).
Step 2: Find the slopes of lines OA and OB
The slope of line OB can be calculated as follows:
slope of OB =
Step 3: Use the angle condition The angle between two lines with slopes m₁ and m₂ is given by:

Given that the angle subtended is π / 4 , we have:

Thus, we can set up the equation:

Step 4: Solve for m₂
This leads to two cases:

Case 1:

Case 2:

Step 5: Find coordinates of points A and A ′
Using the slopes m₂ = 1/ 5 and m₂ = − 5 with the fixed y- coordinate of 2 :

Since the line passes through the origin, b = 0 :


Again, since it passes through the origin, b = 0:

Step 6: Calculate the distance between A and A ′ 
Using the distance formula:

Substituting the coordinates:

Simplifying:

= 52/5

Q.18. The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x − y = 3 respectively. If its ortho centre is , then p is equal to ______________.        (JEE Main 2022)

Ans.3

∴ p² − 8p + 15 = 0
∴ p = 3 or 5
But if p=5 then a=3 not acceptable
∴ p = 3

Q.19. Let be a fixed point in the xy-plane. The image of A in y-axis be B and the image of B in x-axis be C. If D(3 cos⁡θ, a sin⁡θ) is a point in the fourth quadrant such that the maximum area of ΔACD is 12 square units, then a is equal to ____________.      (JEE Main 2022)

Ans.8

Q.20. A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ratio 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (α, β). Then, the value of 7α + 3β is equal to ____________.        (JEE Main 2022)

Ans.31



Bisector of angle PAQ is X = 23/7

⇒ M = (23/7, 8/3)
So, 7α + 3β = 31

Q.21. A man starts walking from the point P(−3, 4), touches the x-axis at R, and then turns to reach at the point Q(0, 2). The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then 50((PR)² + (RQ)²) is equal to ____________.        (JEE Main 2021)

Ans. 1250

To minimize distance PR + RQ

Take mirror image of P in y = 0

P′ = (−3,−4)

If we join P′Q we will get required R

Equation of P′Q 
⇒ y = 2x + 2
So R = (−1, 0)

P = (−3,  4), R(−1, 0), Q(0, 2)
PR² + RQ² = 20 + 5 = 25
50((PR)² + (RQ)²) = 50 x 25 = 1250

Q.22. Let the points of intersections of the lines x − y + 1 = 0, x − 2y + 3 = 0 and 2x − 5y + 11 = 0 are the mid points of the sides of a triangle ΔABC. Then, the area of the ΔABC is _________.        (JEE Main 2021)

Ans.6
Intersection point of given lines are (1, 2), (7, 5), (2, 3)

Q.23. Consider a triangle having vertices A(−2, 3), B(1, 9) and C(3, 8). If a line L passing through the circum-centre of triangle ABC, bisects line BC, and intersects y-axis at point (0, α/2), then the value of real number α is __________.        (JEE Main 2021)

Ans.9

∠B = 90^∘
Circum-center =(1/2, 11/2)
Mid point of BC =(2, 17/2)

Passing through (0, α/2)
α/2 = 9/2 ⇒ α = 9

Q.24. A square ABCD has all its vertices on the curve x²y² = 1. The midpoints of its sides also lie on the same curve. Then, the square of area of ABCD is        (JEE Main 2021)

Ans. 80
x²y² = 1

Graph of this equation,



⇒ p²q² = 1
midpoint of AB lies
On x²y² = 1

⇒ (p + q)²(p − q)² = 16
⇒ (p² − q²)² = 16

⇒ p⁴ ± 4p² − 1 = 0

Q.25. Let tanα, tanβ and be the slopes of three line segments OA, OB and OC, respectively, where O is origin. If circumcentre of ΔABC coincides with origin and its orthocentre lies on y-axis, then the value of is equal to ____________.        (JEE Main 2021)

Ans.144
Since orthocentre and circumcentre both lies on y-axis.
⇒ Centroid also lies on y-axis.

⇒ ∑ cos ⁡α = 0
cos α + cos β + cos γ = 0
⇒ cos³ α + cos³ β + cos³ γ = 3 cos α cos β cos γ

Q.26. The maximum value of z in the following equation z = 6xy + y², where 3x + 4y ≤ 100 and 4x + 3y ≤ 75 for x ≥ 0 and y ≥ 0 is __________.        (JEE Main 2021)

Ans. 904

3x + 4y ≤ 100
4x + 3y ≤ 75
x ≥ 0, y ≥ 0
Feasible region is shown in the graph
Let maximum value of 6xy + y² = c
For a solution with feasible region,
6xy + y² = c and 4x + 3y = 75 must have at least one positive solution.

Q.27. Let λ be an integer. If the shortest distance between the lines x − λ = 2y − 1 = −2z and x = y + 2λ = z − λ is √7/2√2, then the value of | λ | is _________.         (JEE Main 2021)

Ans. 1




Point on line = (0, −2λ, λ)
Distance between skew lines


= |10λ + 3| = 7
⇒ λ = −1

Q.28. Consider the lines L₁ and L₂ defined by L₁ : x√2 + y − 1 = 0 and L₂ : x√2 − y + 1 = 0
For a fixed constant λ, let C be the locus of a point P such that the product of the distance of P from L₁ and the distance of P from L₂ is λ². The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is  Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.  
The value of D is __________.     (JEE Advanced 2021)

Ans. 77.14
According to the question,

Let R ≡ (x₁, y₁) and S(x₂, y₂)
∵ C cuts y − 1 = 2x at R and S.
So, |2x² − 4x²| = 3λ²

∴ |x₁ − x₂| = √6|λ|
and |y₁ − y₂| = 2|x₁ − x₂| = 2√6|λ|
∵ RS² = 270 (given)
⇒ (x₁ − x₂)² + (y₁ − y₂)² = 270
⇒ (√6λ)² + (2√6|λ|)² = 270
⇒ 30λ² = 270 ⇒ λ² = 9
Now, mid-point of RS is  and slope of RS = 2 and slope of R′S′ = −1/2

On solving x + 2y − 2 = 0 with C, we get


Hence, D ≡ (R′S′)² = (x₁ − x₂)² + (y₁ − y₂)²  

Q.29. Consider the lines L₁ and L₂ defined by
L₁ : x√2 + y − 1 = 0 and L₂ : x√2 − y + 1 = 0
For a fixed constant λ, let C be the locus of a point P such that the product of the distance of P from L₁ and the distance of P from L₂ is λ². The line y = 2x + 1 meets C at two points R and S, where the distance between R and S is Let the perpendicular bisector of RS meet C at two distinct points R' and S'. Let D be the square of the distance between R' and S'.
The value of λ² is __________.       (JEE Advanced 2021)

Ans.9
According to the question,

Let R ≡ (x₁, y₁) and S(x₂, y₂)
∵ C cuts y − 1 = 2x at R and S.
So, |2x² − 4x²| = 3λ²



∵ RS² = 270 (given)
⇒(x₁ − x₂)² + (y₁ − y₂)² = 270
⇒(√6λ)² + (2√6|λ|)² = 270
⇒ 30λ² = 270 ⇒ λ² = 9

Q.30. Let A be the set of all points (α, β) such that the area of triangle formed by the points (5, 6), (3, 2) and (α, β) is 12 square units. Then the least possible length of a line segment joining the origin to a point in A, is:         (JEE Main 2021)
(a) 4/√5
(b) 16/√5
(c) 8/√5
(d) 12/√5

Ans.c


4α − 2β = ± 24 + 8
⇒ 4α − 2β = + 24 + 8 ⇒ 2α − β = 16
2x − y − 16 = 0 ..... (1)
⇒ 4α − 2β = − 24 + 8 ⇒ 2α − β = −8
2x − y + 8 = 0 ...... (2)
perpendicular distance of (1) from (0, 0)

perpendicular distance of (2) from (0, 0)

Q.31. If p and q are the lengths of the perpendiculars from the origin on the lines, x cosec α − y sec α = k cot 2α and x sinα + y cosα = k sin 2α respectively, then k² is equal to : (JEE Main 2021) 
(a) 4p² + q²
(b) 2p² + q²
(c) p² + 2q²
(d) p² + 4q

Ans. a
The correct option is a : 4p² + q²
Given lines are
xcosecα − ysecα = kcot2αand xsinα + ycosα = ksin2α
Distance from origin are

And

⇒ q² = k²sin²2α⋯(2)
Adding equation (1)and (2), we get
4p² + q² = k²

Q.32. The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m − n = 0 and mn + nl + lm = 0, is :        (JEE Main 2021) 
(a) π/2
(b)
(c)
(d) π/3

Ans.a

n = 2 (l + m)
lm + n(l + m) = 0
lm + 2(l + m)² = 0
2l² + 2m² + 5ml = 0
2(l/m)² + 2 + 5(l/m) = 0
2t² + 5t + 2 = 0
(t + 2)(2t + 1) = 0

(i) l/m = −2
(−2m, m, −2m)
(−2, 1, −2)

n = −2l
(l, −2l, −2l)
(1, −2, −2)

Q.33. Let A be a fixed point (0, 6) and B be a moving point (2t, 0). Let M be the mid-point of AB and the perpendicular bisector of AB meets the y-axis at C. The locus of the mid-point P of MC is :        (JEE Main 2021) 
(a) 3x² − 2y − 6 = 0
(b) 3x² + 2y − 6 = 0
(c) 2x² + 3y − 9 = 0
(d) 2x² − 3y + 9 = 0

Ans.c

A(0, 6) and B(2t, 0)

Perpendicular bisector of AB is


Let P be (h, k)

Q.34. Let ABC be a triangle with A(−3, 1) and ∠ACB = θ, 0 < θ < π/2. If the equation of the median through B is 2x + y − 3 = 0 and the equation of angle bisector of C is 7x − 4y − 1 = 0, then tanθ is equal to:        (JEE Main 2021)
(a) 1/2
(b) 3/4
(c) 4/3
(c) 2

Ans. c

lies on 2x + y − 3 = 0
⇒ 2a + b = 11 ...........(i)
∵ C lies on 7x − 4y = 1
⇒ 7a − 4b = 1 ......... (ii)
∴ by (i) and (ii) : a = 3, b = 5
⇒ C(3, 5)
∴ m_AC = 2/3
Also, m_CD = 7/4

Q.35. Let the equation of the pair of lines, y = px and y = qx, can be written as (y − px) (y − qx) = 0. Then the equation of the pair of the angle bisectors of the lines x² − 4xy − 5y² = 0 is : (JEE Main 2021) 
(a) x² − 3xy + y² = 0
(b) x² + 4xy − y² = 0
(c) x² + 3xy − y² = 0
(d) x² − 3xy − y² = 0

Ans.c
Equation of angle bisector of homogeneous equation of pair of straight line ax² + 2hxy + by² is

for x² – 4xy – 5y² = 0
a = 1, h = – 2, b = – 5
So, equation of angle bisector is


⇒ x² − y² = −3xy
So, combined equation of angle bisector is x² + 3xy - y² = 0

Q.36. Two sides of a parallelogram are along the lines 4x + 5y = 0 and 7x + 2y = 0. If the equation of one of the diagonals of the parallelogram is 11x + 7y = 9, then other diagonal passes through the point :        (JEE Main 2021) 
(a) (1, 2)
(b) (2, 2)
(c) (2, 1)
(d) (1, 3)

Ans.b
Both the lines pass through origin.

point D is equal to intersection of 4x + 5y = 0 & 11x + 7y = 9
So, coordinates of point
Also, point B is point of intersection of 7x + 2y = 0 and 11x + 7y = 9
So, coordinates of point
diagonals of parallelogram intersect at middle let middle point of B, D

equation of diagonal AC

y = x
diagonal AC passes through (2, 2)

Q.37. The point P (a, b) undergoes the following three transformations successively :
(A) reflection about the line y = x.
(B) translation through 2 units along the positive direction of x-axis.
(C) rotation through angle π/4 about the origin in the anti-clockwise direction.
If the co-ordinates of the final position of the point P are , then the value of 2a + b is equal to : (JEE Main 2021) 
(a) 13
(b) 9
(c) 5
(d) 7

Ans.b
Image of A(a, b) along y = x is B(b, a). Translating it 2 units it becomes C(b + 2, a).
Now, applying rotation theorem

⇒ b − a + 2 = −1 ......(i)
and b + 2 + a = 7 ...... (ii)
⇒ a = 4; b = 1
⇒ 2a + b = 9

Q.38. Let the centroid of an equilateral triangle ABC be at the origin. Let one of the sides of the equilateral triangle be along the straight line x + y = 3. If R and r be the radius of circumcircle and incircle respectively of ΔABC, then (R + r) is equal to :        (JEE Main 2021) 
(a) 7√2
(b) 9/√2
(c) 2√2
(d) 3√2

Ans. b

Finding the value of (R + r):

Illustrating a figure according to given data

Given that one of the sides of the equilateral triangle be along the straight line 
x + y = 3
And we know in an equilateral triangle each side have an angle $60°$

So, considering the given line bisects the angle.

Therefore,

So, from the figure

Hence, option (b) is the correct answer.

Q.39. The equation of one of the straight lines which passes through the point (1, 3) and makes an angles tan⁻¹(√2) with the straight line, y + 1 = 3√2 x is        (JEE Main 2021)
(a) 4√2x + 5y−(15 + 4√2) = 0
(b) 5√2x + 4y − (15 + 4√2) = 0
(c) 4√2x + 5y − 4√2 = 0
(d) 4√2x − 5y − (5 + 4√2) = 0

Ans.a
Let slope of line be m




Hence line can be

Q.40. The number of integral values of m so that the abscissa of point of intersection of lines 3x + 4y = 9 and y = mx + 1 is also an integer, is :        (JEE Main 2021)
(a) 1
(b) 2
(c) 3
(d) 0

Ans.b
3x + 4(mx + 1) = 9
⇒ x(3 + 4m) = 5

⇒ (3 + 4m) = ±1, ±5
⇒ 4m = −3 ± 1, −3 ± 5
⇒ 4m = −4, −2, −8, 2

∴ Two integral value of m.

Q.41. In a triangle PQR, the co-ordinates of the points P and Q are (−2, 4) and (4, −2) respectively. If the equation of the perpendicular bisector of PR is 2x − y + 2 = 0, then the centre of the circumcircle of the ΔPQR is :        (JEE Main 2021) 
(a) (−1, 0)
(b) (1, 4)
(c) (0, 2)
(d) (−2, −2)

Ans.d


Slope of perpendicular bisector of PQ = 1
Equation of perpendicular bisector of PQ
y − 1 = 1(x − 1)
⇒ y = x
Solving with perpendicular bisector of PR,
2x − y + 2 = 0
Circumcentre is (−2, −2)

Q.42. Let A(−1, 1), B(3, 4) and C(2, 0) be given three points.
A line y = mx, m > 0, intersects lines AC and BC at point P and Q respectively. Let A₁ and A₂ be the areas of ΔABC and ΔPQC respectively, such that A₁ = 3A₂, then the value of m is equal to :        (JEE Main 2021)
(a) 1
(b) 3
(c) 2
(d) 415

Ans.a


A₁ = 13/2
Equation of line AC is
Line AC intersect with line y = mx at P,
Solving we get
Equation of line BC is y − 0 = 4(x − 2)
Line BC intersect with line y = mx at Q,
Solving we get


⇒12m² = ±(3m² −11m − 4)
taking +ve sign
9m² + 11m + 4 = 0 (Rejected ∵ m is imaginary)
taking −ve sign
15m² − 11m − 4 = 0

∴ m = 1 (As given m > 0)

Q.43. The intersection of three lines x − y = 0, x + 2y = 3 and 2x + y = 6 is a :        (JEE Main 2021) 
(a) Right angled triangle
(b) Equilateral triangle
(c) None of the above
(d) Isosceles triangle

Ans.d

The given three lines are x − y = 0, x + 2y = 3 and 2x + y = 6 then point of intersection,
lines x − y = 0 and x + 2y = 3 is (1, 1)
lines x − y = 0 and 2x + y = 6 is (2, 2)
and lines x + 2y = 3 and 2x + y = 0 is (3, 0)
The triangle ABC has vertices A(1, 1), B(2, 2) and C(3, 0)
∴ AB = √2, BC = √5 and AC = √5
∴ ΔABC is isosceles

Q.44. The image of the point (3, 5) in the line x − y + 1 = 0, lies on :        (JEE Main 2021) 
(a) (x − 4)² + (y − 4)² = 8
(b) (x − 4)² + (y + 2)² = 16
(c) (x − 2)² + (y − 2)² = 12
(d) (x − 2)² + (y − 4)² = 4

Ans.d
So, let the image is (x, y)
So, we have

⇒ x = 4, y = 4
⇒ Point (4, 4)
Which will satisfy the curve
(x − 2)² + (y − 4)² = 4
as (4 − 2)² + (4 − 4)² = 4 + 0 = 4

Q.45. A man is walking on a straight line. The arithmetic mean of the reciprocals of the intercepts 14 of this line on the coordinate axes is Three stones A, B and C are placed at the points (1, 1), (2, 2) and (4, 4) respectively. Then which of these stones is/are on the path of the man ?        (JEE Main 2021) 
(a) A only
(b) All the three
(c) C only
(d) B only

Ans.d
Given, position of A = (1, 1)
Position of B = (2, 2)
Position of C = (4, 4)


Let x-intercept be a and y-intercept be b.
Equation of line traced is

This is the equation of path, let a point (h, k) lie on this path.

Also, AM of reciprocal of a and b = 1/4


On comparing Eqs. (i) and (ii), we get (h, k) = (2, 2)
Hence, the required stone is B(2, 2).

Q.46. Let A(1 , 0), B(6, 2) and 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 the length of the line segment PQ, where Q is the point (-7/6, -1/3), is ___.    (2020)

Ans. (5.00)
Finding the length of line segment$\mathrm{ PQ}$

Step 1: Determine the value of$\mathrm{ P}$

Given the vertices of a triangle ABC are A$(1,0)$, B$(6,2)$and C(3/2, 6)

P is a point inside the given triangle such that the triangles APC, APB and BPC have equal area, therefore the point P will be the centroid of a triangle ABC.

Point Qhas coordinates (-7/6, -1/3)

The centroid of a triangle is calculated by adding the vertices of a triangle and dividing by 3. Hence,

Step 2: Determine the Length ofPQ

Now the length of a line segment PQ will be:

Substituting the values we have:

Hence, the length of a line segment PQ is equal to 5.

Q.47. The locus of the mid-points of the perpendicular drawn from points on the line x = 2y to the line x = y is    (2020)
(a) 2x – 3y = 0
(b) 5x – 7y = 0
(c) 3x – 2y = 0
(d) 7x – 5y = 0

Ans. b
Let the mid-point of perpendicular is (h, k).

Now, the slope of perpendicular lines is

Now,   ...(1)
...(2)
From Eqs. (1) and (2), we get

Hence, the locus of mid-point of perpendiculars is
5x = 7y ⇒ 5x - 7y = 0.

Q.48. Let two points be A (1, −1) and B (0, 2). If a point P(x', y')be such that the area of ΔPAB = 5 sq. units and it lies on the line 3x + y - 4λ = 0, then a value of λ is    (2020)
(a) 4
(b) 3
(c) 2
(d) −3

Ans. b
The equation of the line passing through the points A (1, −1) and B (0, 2) is
...(1)
Now,

The lines 3x + 2y - 2 = 0 and 3x + y - 4λ = 0 are parallel lines and distance between them is equal to the height of triangle PAB. So,
...(2)
The area of the triangle PAB is 5 sq. units, then

⇒ λ = -2, 3

Q.49. Let C be the centroid of the triangle with 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 passes through the point:    (2020)
(a) (−9, −6)
(b) (9, 7)
(c) (7, 6)
(d) (−9, −7)

Ans. a
The coordinates of point C is

The intersection point P of the lines x + 3y - 1 = 0 and 3x - y  + 1 = 0 is
x = -1/5 and y = 2/5
Now, the equation of line CP is

⇒ 8x - 11y + 6 = 0
Hence, from the given options , the line passes through the point (−9, −6).
⇒ 8(-9) + 66 + 6 = 0

Q.50. Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true?    (2019)
(a)The lines are concurrent at the point
(b)Each line passes through the origin.
(c)The lines are all parallel.
(d)The lines are not concurrent.

Ans. a
The given equations of the set of all lines
px + qy + r = 0   ...(1)
and given condition is:
3p + 2q + 4r = 0
...(2)
From (1) & (2) we get:

Hence the set of lines are concurrent and passing through the fixed point

Q.51. Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is: (2019)
(a) 9   
(b) 18
(c) 36   
(d) 32

Ans. c
One of the possible ΔOAB is A(a, 0) and B(0, b).


no. of integral solutions of the equation is the number of
divisors of 100 : 1, 2, 4, 5, 10, 20, 25, 50, 100 = Total divisors = 9  
So, (9 × 4) = 36

Q.52. Let the equations of two sides of a triangle be 3x - 2y + 6 = 0 and 4x + 5y - 20 = 0. If the orthocentre of this triangle is at (1, 1), then the equation of its third side is:    (2019)
(a) 122y - 26x - 1675 = 0
(b)122y + 26x + 1675 = 0
(c)26x + 61y + 1675 = 0
(d)26x - 122y - 1675 = 0

Ans. d

Since, AH is perpendicular to BC

Q.53. A point P moves on the line 2x - 3y + 4 = 0. If Q(1, 4) and R(3, -2) are fixed points, then the locus of the centroid of ΔPQR is a line:    (2019)
(a)with slope 3/2
(b)parallel to x-axis
(c) with slope 2/3
(d)parallel to y-axis

Ans. c

Q.54. If the line 3x + 4y - 24 = 0 intersects the x-axis at the point A and the x-axis at the point B, then the incentre of the triangle OAB, where O is the origin, is:    (2019)
(a) (3,4)   
(b) (2,2)
(c) (4,3)   
(d) (4,4)

Ans. b
Equation of the line is:

Q.55. Two sides of a parallelogram are along the lines, x + y = 3 and x - y + 3 = 0. If its diagonals intersect at (2, 4), then one of its vertex is:    (2019)
(a) (3,5)   
(b) (2,1)
(c) (2,6)   
(d) (3,6)

Ans.d

Since, x - y + 3 = 0 and x + y = 3 are perpendicular lines and intersection point of x - y + 3 = 0 and x + y = 3 is P(0,3).

Then, the vertex D is (3, 6).

Q.56. Two vertices of a triangle are (0, 2) and (4, 3). If its orthocentre is at the origin, then its third vertex lies in which quadrant?    (2019)
(a) third   
(b) second
(c) first   
(d) fourth

Ans.b


Hence, vertex R lies in second quadrant.

Q.57. The straight line x + 2y = 1 meets the coordinate axes at A and B. A circle is drawn through A, B and the origin. Then the sum of perpendicular distances from A and B on the tangent to the circle at the origin is:    (2019)
(a)    
(b)
(c) 
(d)

Ans. a
Equation of circle

Q.58. If in a parallelogram ABDC, the coordinates of A, B and C are respectively (1, 2), (3,4) and (2,5), then the equation of the diagonal AD is:    (2019)
(a) 5x - 3y + 1=0   
(b) 5x + 3y - 11=0
(c) 3x - 5y + 7 = 0   
(d) 3x + 5y - 13 = 0

Ans.a
Since, in parallelogram mid points of both diagonals coincides.
∴ mid-point of AD = mid-point of BC

=

Q.59. If the straight line, 2x - 3y + 17 = 0 is perpendicular to the line passing through the points (7, 17) and (15, β), then β equals:    (2019)
(a) 35/3   
(b) -5
(c) -35/3
(d) 5

Ans. d
∵ Equation of straight line can be rewritten as,

∴ Slope of straight line = 2/3
Slope of line passing through the points (7, 17) and

Since, lines are perpendicular to each other.
Hence, m₁m₂ = -1

Q.60. If a straight line passing through the point P(-3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is:    (2019)
(a) 3x - 4y + 25 = 0   
(b) 4x - 3y + 24 = 0
(c) x - y + 7 = 0   
(d) 4x + 3y = 0

Ans.b
Since, P is mid point of MN


⇒ y = 8
Hence required equation of straight line MN is

⇒ 4x - 3y + 24 = 0

Q.61. A point on the straight line, 3x + 5y =15 which is equidistant from the coordinate axes will lie only in:    (2019)
(a)4^th quadrant
(b)1^st quadrant
(c)1^st and 2^nd quadrants
(d)1^st, 2^nd and 4^th quadrants

Ans. c
A point which is equidistant from both the axes lies on either y = x and y = -x.
Since, point lies on the line 3x + 5y = 15
Then the required point
3x + 5y = 15


Hence, the required point lies in 1^st and 2^nd quadrant.

Q.62. Let 0(0, 0) and A(0, 1) be two fixed points. Then the locus of a point P such that the perimeter of ΔAOP is 4, is:    (2019)
(a)8x² - 9y² + 9y = 18
(b)9x² - 8y²+ 8y = 16
(c)9x² + 8y² - 8y = 16
(d)8x² + 9y² - 9y = 18

Ans. c
Let point P (h, k)
∵ OA = 1
So, OP + AP = 3

Q.63. Suppose that the points (h, k), (1, 2) and (-3, 4) lie on the line L₁. If a line L₂ passing through the points (h, k) and (4, 3) is perpendicular on L₁, then k/h equals: (2019)
(a) 1/3   
(b) 0   
(c) 3   
(d) -1/7

Ans. a
∵ (h, k), (1,2) and (- 3, 4) are collinear

⇒ h + 2k = 5   ...(i)

By the given points (h, k) and (4, 3),

Q.64. Slope of a line passing through P(2, 3) and intersecting the line x + y = 7 at a distance of 4 units from P, is:    (2019)
(a)
(b)
(c)
(d)

Ans. b

Since point at 4 units from P (2, 3) will be
A (4 cosθ + 2, 4 sin (θ + 3) and this point will satisfy the equation of line x + y = 7

Q.65. If the two lines x + (a - 1)y = 1 and 2x + a²y =1 (a ∈ R - {0,1}) are perpendicular, then the distance of their point of intersection from the origin is: (2019)
(a)
(b) 2/5
(c)
(d)

Ans. a
∵ two lines are perpendicular ⇒ m₁m₂ = - 1

Hence equations of lines are x - 2y= 1 and 2x + y = 1
∴ intersection point is

Now, distance from origin

Q.66. A rectangle is inscribed in a circle with a diameter lying along the line 3y = x + 7. If the two adjacent vertices of the rectangle are (-8, 5) and (6, 5), then the area of the rectangle (in sq. units) is:    (2019)
(a) 84   
(b) 98   
(c) 72   
(d) 56

Ans. a
Given situation

∵ perpendicular bisector of AB will pass from centre.
∴ equation of perpendicular bisector x = -1
Hence centre of the circle is (- 1, 2)
Let co-ordinate of D is (α, β)

⇒ α = - 8 and β = - 1, D ≡ (-8, -1)
|AD| = 6 and |AB| = 14
Area = 6 x 14 = 84 sq.units

Q.67. Lines are drawn parallel to the line 4x - 3y + 2 = 0, at a distance 3/5 from the origin. Then which one of the following points lies on any of these lines ?    (2019)
(a)
(b)
(c)
(d)

Ans.a
Let straight line be 4x - 3y + α = 0
∵ distance from origin = 3/5

Hence, line is 4x - 3y + 3 = 0 or 4x - 3y - 3 = 0

Q.68. The equation y = sin x sin (x + 2) - sin² (x + 1) represents a straight line lying in: (2019)
(a)second and third quadrants only
(b)first, second and fourth quadrant
(c)first, third and fourth quadrants
(d)third and fourth quadrants only

Ans.d
Consider the equation,
y = sin x. sin (x + 2) - sin²(x + 1)


By the graph y lies in III and IV quadrant.

Q.69. A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3). Then the centroid of this triangle is:    (2019)
(a)
(b)
(c)
(d)

Ans. b

Q.70. A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the line x + y = 0. Then an equation of the line L is:    (2019)
(a)
(b)
(c)
(d)

Ans. b, d
∵ Perpendicular makes an angle of 60° with the line x + y = 0.
∴ The perpendicular makes an angle of 15° or 75° with x-axis.

Hence, the equation of line will be
x cos 75° + y sin 75° = 4
or x cos 15° +y sin 15° = 4

Q.71. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:    (2018)
(a) 3x + 2y = 6
(b) 2y + 3x = xy
(c) 3x + 2y = xy
(d) 3x + 2y = 6xy

Ans. b
Let R = (h,k)
P = (0, k)
Q = (h,0)
Equation of line would be,


2k + 3h = hk
Locus of (h, k) is 2y + 3x = xy

Q.72. Let k be an integer such that the triangle with vertices (k, –3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point    (2017)
(a)
(b)
(c)
(d)

Ans. a
We have

or
5k² + 13k -46 = 0 (no real solution exist)
∴ k = -23/5
or k = 2
k is an integer, so k=2

Therefore (2, 1/2)

Q.73. A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 30° with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is:    (2017)
(a) 2√3-2
(b) √3-2
(c) 2√3-1
(d) √3-1

Ans. a

Q.74. Two sides of a rhombus are along the lines, x - y + 1 = 0 and 7x - y - 5 = 0. If its diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus?    (2016)
(a) (-3, -9)
(b) (-3, -8)
(c)
(d)

Ans. c
Coordinates of A = (1, 2)
∴ Slope of AE = 2

⇒ Slope of BD =
⇒ Eq. of BD is
⇒ x + 2y + 5 = 0
Co-ordinates of D =

Q.75. If a variable line drawn through the intersection of the lines and meets the coordinate axes at A and B, (A ≠ B), then the locus of the midpoint of AB is (2016)
(a) 7xy = 6(x + y) 
(b) 6xy = 7(x + y) 
(c) 4(x + y)² – 28(x + y) + 49 = 0 
(d) 14(x + y)² – 97 (x + y) + 168 = 0

Ans.a
4x + 3y = 12 ....(1)
3x + 4y = 12 ....(2)
equation of lines passing through the intersection of the lines

4x + 3y – 12 + λ(3x + 4y - 12) = 0


Hence, locus is 7xy = 6(x + y)

Q.76. The point (2, 1) is translated parallel to the line L : x – y = 4 by units. If the new point Q lies in the third quadrant, then the equation of the line passing through Q and perpendicular to L is:    (2016)
(a) 2x + 2y = 1 - √6
(b) x = y = 3 - 3 √6
(c) x + y = 2 - √6
(d) x + y =  3 - 2 √6

Ans. d
Slopes of x – y = 4

⇒ Equation of required line is

Q.77. A straight line through origin O meets the line 3y = 10 – 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio    (2016)
(d) 3 : 4
(b) 1 : 2
(c) 2 : 3
(d) 4 : 1

Ans. d

Q.78. A ray of light is incident along a line which meets another line, 7x – y + 1 = 0, at the point (0, 1). The ray is then reflected from this point along the line, y + 2x = 1. Then the equation of the line of incidence of the ray of light is    (2016)
(a) 41x + 38y – 38 = 0 
(b) 41x – 38y + 38 = 0 
(c) 41x + 25y – 25 = 0 
(d) 41x – 25y + 25 = 0

Ans. b
Incident line


Let a point (1, –1) on  y + 2x = 1
And image of (1, –1) lie on incident line in
7x – y + 1 = 0

⇒ 369x –342y + 342 = 0
⇒ 41x –38y + 38 = 0