Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2 and x + y = 2 , respectively, and the…

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More Straight Lines Previous-Year Questions — Page 6

Q1 jee_main_2024_30_jan_morning Rotation of Axes and Lines

A line passing through the point

A(9,0)

makes an angle of

30°

with the positive direction of x-axis. If this line is rotated about A through an angle of

15°

in the clockwise direction, then its equation in the new position is

A. $y\sqrt(3) - 2 + x = 9$ $\frac{y}{\sqrt{3} - 2} + x = 9$
B. $x\sqrt(3) - 2 + y = 9$ $\frac{x}{\sqrt{3} - 2} + y = 9$
C. $x\sqrt(3) + 2 + y = 9$ $\frac{x}{\sqrt{3} + 2} + y = 9$
D. $y\sqrt(3) + 2 + x = 9$ $\frac{y}{\sqrt{3} + 2} + x = 9$

Solution

Related Formula

y - y₁ = (θ)(x - x₁)

Core Logic

Rotation of Axes and Lines diagram for Q1 - JEE Main 2024 Morning

The initial line makes an angle of $30°$ $30^{\circ}$ with the positive x-axis. It is rotated clockwise by $15°$ $15^{\circ}$ about the point $A(9, 0)$ $A(9, 0)$. The new angle made by the line with the positive direction of the x-axis is $30° - 15° = 15°$ $30^{\circ} - 15^{\circ} = 15^{\circ}$.

Step 1: Equation of the new line

The equation of the line passing through $A(9, 0)$ $A(9, 0)$ with a slope of $15°$ $\tan 15^{\circ}$ is:

Eqⁿ: y - 0 = 15° (x - 9)

We know that $15° = 2 - \sqrt(3)$ $\tan 15^{\circ} = 2 - \sqrt{3}$.

y = (2 - \sqrt(3)) (x - 9)

Step 2: Rearranging to match options

Dividing by $(2 - \sqrt(3))$ $(2 - \sqrt{3})$:

y2 - \sqrt(3) = x - 9

Notice that $2 - \sqrt(3) = -(\sqrt(3) - 2)$ $2 - \sqrt{3} = -(\sqrt{3} - 2)$. Thus:

-y\sqrt(3) - 2 = x - 9

y\sqrt(3) - 2 + x = 9

Pattern Recognition

A clockwise rotation decreases the angle of inclination. Calculate the new angle, find its tangent, and carefully algebraicize the denominator to match the given option forms.

Chapter Mix

Class 11 Maths: Straight Lines

Q2 jee_main_2024_31_jan_evening Centroid and Orthocentre

Let

A (a, b)

B(3, 4)

and

(-6, -8)

respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point

P(2a + 3, 7b + 5)

from the line

2x + 3y - 4 = 0

measured parallel to the line

x - 2y - 1 = 0

A. $15 \sqrt(5)7$ $\frac{15 \sqrt{5}}{7}$
B. $17\sqrt(5)6$ $\frac{17\sqrt{5}}{6}$
C. $17 \sqrt(5)7$ $\frac{17 \sqrt{5}}{7}$
D. $\sqrt(5)17$ $\frac{\sqrt{5}}{17}$

Solution

Related Formula

Centroid divides the line joining Orthocentre and Circumcentre in 2:1

Distance in parametric form: $x = x₁ + r θ, y = y₁ + r θ$

Core Logic

Centroid and Orthocentre diagram for Q2 - JEE Main 2024 Evening

Let Orthocentre $C(-6, -8)$ $C(-6, -8)$ and Circumcentre $B(3, 4)$ $B(3, 4)$. Centroid $A(a, b)$ $A(a, b)$ divides $CB$ $CB$ in $2:1$ $2:1$.

a = (2(3) + 1(-6))/(2+1) = 0

b = (2(4) + 1(-8))/(2+1) = 0

So, $P(2a+3, 7b+5) = (3, 5)$ $P(2a+3, 7b+5) = (3, 5)$.

The line along which distance is measured is parallel to $x - 2y - 1 = 0$ $x - 2y - 1 = 0$, giving slope $m = θ = (1)/(2)$ $m = \tan\theta = \frac{1}{2}$. Using parametric coordinates from $P(3,5)$ $P(3,5)$:

x = 3 + r θ, y = 5 + r θ

Substitute into the target line $2x + 3y - 4 = 0$ $2x + 3y - 4 = 0$:

2(3 + r θ) + 3(5 + r θ) - 4 = 0

r(2 θ + 3 θ) = -17

From $θ = 1/2$ $\tan\theta = 1/2$, we get $θ = 1\sqrt(5)$ $\sin\theta = \frac{1}{\sqrt{5}}$ and $θ = 2\sqrt(5)$ $\cos\theta = \frac{2}{\sqrt{5}}$.

r(2( 2\sqrt(5)) + 3( 1\sqrt(5))) = -17

r( 7\sqrt(5)) = -17 |r| = 17\sqrt(5)7

Pattern Recognition

Standard Euler line property: $O, G, C$ $O, G, C$ are collinear and $G$ $G$ divides $OC$ $OC$ in $2:1$ $2:1$. Use parametric equation to find intersection distance directly without finding the intersection point.

Chapter Mix

Class 11 Maths: Straight Lines

Q23 jee_main_2024_31_jan_evening Parallelogram Properties

Let

A(-2, -1)

B(1, 0)

C(α, β)

and

D(γ, δ)

be the vertices of a parallelogram ABCD. If the point C lies on

2x - y = 5

and the point D lies on

3x - 2y = 6

, then the value of

|α + β + γ + δ|

is equal to

Numerical Answer. Answer: 32 to 32

Solution

Related Formula

In a parallelogram, midpoints of diagonals coincide: ((xA+xC)/(2), (yA+yC)/(2)) = ((xB+xD)/(2), (yB+yD)/(2))

Core Logic

Parallelogram Properties diagram for Q23 - JEE Main 2024 Evening

Given diagonals $AC$ $AC$ and $BD$ $BD$ bisect each other:

α - 2 = γ + 1 α - γ = 3 (1)

β - 1 = δ + 0 β - δ = 1 (2)

Point $C(α, β)$ $C(\alpha, \beta)$ lies on $2x - y = 5 2α - β = 5 (3)$ Point $D(γ, δ)$ $D(\gamma, \delta)$ lies on $3x - 2y = 6 3γ - 2δ = 6 (4)$

From (1) and (2), substitute $γ = α - 3$ $\gamma = \alpha - 3$ and $δ = β - 1$ $\delta = \beta - 1$ into (4):

3(α - 3) - 2(β - 1) = 6

3α - 9 - 2β + 2 = 6 3α - 2β = 13 (5)

Solve (3) and (5): From (3), $β = 2α - 5$ $\beta = 2\alpha - 5$. Substitute in (5):

3α - 2(2α - 5) = 13

-α + 10 = 13 α = -3

So, $β = 2(-3) - 5 = -11$ $\beta = 2(-3) - 5 = -11$. From earlier substitutions:

γ = -3 - 3 = -6

δ = -11 - 1 = -12

Sum of variables:

|α + β + γ + δ| = |-3 - 11 - 6 - 12| = |-32| = 32

Chapter Mix

Class 11 Maths: Straight Lines

Q10 jee_main_2024_31_jan_morning Properties of Parallelogram

Let

α, β, γ, δ in Z

and let

A (α, β)

B (1, 0)

C (γ, δ)

and

D (1, 2)

be the vertices of a parallelogram

ABCD

. If

AB = \sqrt(10)

and the points

A

and

C

lie on the line

3y = 2x + 1

, then

2(α + β + γ + δ)

is equal to

A. $10$ $10$
B. $5$ $5$
C. $12$ $12$
D. $8$ $8$

Solution

Core Logic

Let $E$ $E$ be the midpoint of the diagonals $AC$ $AC$ and $BD$ $BD$. Since $ABCD$ $ABCD$ is a parallelogram, the diagonals bisect each other.

Properties of Parallelogram diagram for Q10 - JEE Main 2024 Morning

Midpoint

E

from

BD

( (1+1)/(2), (0+2)/(2) ) = (1, 1)

Step 1: Apply Midpoint on AC

Midpoint $E$ $E$ from $AC$ $AC$: $( (α+γ)/(2), (β+δ)/(2) )$ . Equating both:

(α+γ)/(2) = 1 α + γ = 2

(β+δ)/(2) = 1 β + δ = 2

Step 2: Final Value

The expression requires $2(α + β + γ + δ)$ $2(\alpha + \beta + \gamma + \delta)$.

2(2 + 2) = 2(4) = 8

(Note: Additional conditions like $AB = \sqrt(10)$ $AB = \sqrt{10}$ and the line equation are extraneous data not needed to find the sum).

Chapter Mix

Class 11 Maths: Straight Lines

Let ABC be the triangle such that the equations of lines AB and AC be 3y - x = 2$3y - x = 2$ and x + y = 2$x + y = 2$ , respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Find Equations of Altitudes

Step 2: Solve for Orthocentre coordinates P

Step 3: Compute Area of Triangle PBC

Pattern Recognition

Chapter Mix

Reference Study Guides

More Straight Lines Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Equation of the new line

Step 2: Rearranging to match options

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Apply Midpoint on AC

Step 2: Final Value

Chapter Mix

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Let ABC be the triangle such that the equations of lines AB and AC be $3y - x = 2$ $3y - x = 2$ and $x + y = 2$ $x + y = 2$ , respectively, and the points B and C lie on x-axis. If P is the orthocentre of the triangle ABC, then the area of the triangle PBC is equal to