In a Delta ABC$\Delta ABC$, suppose y=x$y=x$ is the equation of the bisector of the angle B and the equation of the side AC is 2x-y=2$2x-y=2$. If 2AB=BC$2AB=BC$ and the point A and B are respectively (4,6)$(4,6)$ and (alpha,beta)$(\alpha,\beta)$, then alpha+2beta$\alpha+2\beta$ is equal to
A.42$42$
B.39$39$
C.48$48$
D.45$45$
Solution & Explanation
### Related Formula
Image of a point (x_1, y_1)$(x_1, y_1)$ across line y=x$y=x$ is (y_1, x_1)$(y_1, x_1)$.
**Angle Bisector Theorem:** The angle bisector of a triangle divides the opposite side into segments proportional to the lengths of the adjacent sides:
fracABBC = fracADDC$$\frac{AB}{BC} = \frac{AD}{DC}$$
### Core Logic
Given A(4,6)$A(4,6)$ and Angle bisector of B is y=x$y=x$.
Because y=x$y=x$ bisects angle B, the geometric reflection of vertex A across the bisector line y=x$y=x$ must lie exactly on the line containing the side BC$BC$.
Let the reflection of A(4,6)$A(4,6)$ be A'$A'$. Across y=x$y=x$, the coordinates swap:
A' = (6,4)$A' = (6,4)$
Next, find the intersection point D$D$ of the bisector y=x$y=x$ and side AC$AC$ (2x-y=2$2x-y=2$).
Substitute y=x$y=x$ into 2x-y=2$2x-y=2$:
2x - x = 2 Rightarrow x = 2 Rightarrow y = 2$$2x - x = 2 \Rightarrow x = 2 \Rightarrow y = 2$$
So, point D$D$ is (2,2)$(2,2)$.
Angle Bisector and Reflection
### Step 1: Utilize Section Formula
By the internal angle bisector theorem:
fracADDC = fracABBC$$\frac{AD}{DC} = \frac{AB}{BC}$$
Given 2AB = BC$2AB = BC$, so fracABBC = frac12$\frac{AB}{BC} = \frac{1}{2}$.
This means point D(2,2)$D(2,2)$ divides the segment AC$AC$ in the ratio 1:2$1:2$.
Let C$C$ have coordinates (x_c, y_c)$(x_c, y_c)$.
Applying the section formula for D(2,2)$D(2,2)$ dividing A(4,6)$A(4,6)$ and C(x_c, y_c)$C(x_c, y_c)$ in ratio 1:2$1:2$:
2 = frac1 cdot x_c + 2 cdot 41 + 2 Rightarrow 6 = x_c + 8 Rightarrow x_c = -2$$2 = \frac{1 \cdot x_c + 2 \cdot 4}{1 + 2} \Rightarrow 6 = x_c + 8 \Rightarrow x_c = -2$$2 = frac1 cdot y_c + 2 cdot 61 + 2 Rightarrow 6 = y_c + 12 Rightarrow y_c = -6$$2 = \frac{1 \cdot y_c + 2 \cdot 6}{1 + 2} \Rightarrow 6 = y_c + 12 \Rightarrow y_c = -6$$
So, C$C$ is (-2,-6)$(-2,-6)$.
### Step 2: Find Equation of BC
The line BC$BC$ passes through point C(-2,-6)$C(-2,-6)$ and the reflection point A'(6,4)$A'(6,4)$.
Find the slope of BC$BC$:
m_BC = frac4 - (-6)6 - (-2) = frac108 = frac54$$m_{BC} = \frac{4 - (-6)}{6 - (-2)} = \frac{10}{8} = \frac{5}{4}$$
Equation of BC$BC$:
y - 4 = frac54(x - 6)$$y - 4 = \frac{5}{4}(x - 6)$$4y - 16 = 5x - 30$$4y - 16 = 5x - 30$$5x - 4y - 14 = 0$$5x - 4y - 14 = 0$$
### Step 3: Solve for Vertex B
Vertex B(alpha, beta)$B(\alpha, \beta)$ is the intersection of line BC$BC$ and the angle bisector y=x$y=x$.
Substitute y=x$y=x$ into 5x - 4y - 14 = 0$5x - 4y - 14 = 0$:
5x - 4x - 14 = 0 Rightarrow x = 14$$5x - 4x - 14 = 0 \Rightarrow x = 14$$
Thus, y = 14$y = 14$.
Therefore, B$B$ is (14, 14)$(14, 14)$, implying alpha = 14$\alpha = 14$ and beta = 14$\beta = 14$.
Calculate alpha + 2beta$\alpha + 2\beta$:
alpha + 2beta = 14 + 2(14) = 42$$\alpha + 2\beta = 14 + 2(14) = 42$$
### Pattern Recognition
Reflection properties drastically simplify angle bisector questions. If you know the bisector equation, reflecting one vertex over it gives a coordinate on the opposing extended ray. This paired with the angle bisector proportion theorem locks the entire geometric frame.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Keywords:#equation of the bisector#JEE Main 2024 Morning Q7#Straight Lines JEE Main 2024#Angle Bisector and Reflection JEE Main 2024
More Straight Lines Previous-Year Questions
Q4jee_main_2026_21_jan_morningEquilateral Triangle Between Parallel Lines
Let a point A lie between the parallel lines L_1$L_{1}$ and L_2$L_{2}$ such that its distances from L_1$L_{1}$ and L_2$L_{2}$ are 6 and 3 units, respectively. Then the area (in sq. units) of the equilateral triangle ABC, where the points B and C lie on the lines L_1$L_{1}$ and L_2$L_{2}$ respectively, is:
A.15sqrt6$15\sqrt{6}$
B. 27
C.21sqrt3$21\sqrt{3}$
D.12sqrt2$12\sqrt{2}$
Solution
### Related Formula
textArea of Equilateral Triangle = fracsqrt34 a^2$$\text{Area of Equilateral Triangle} = \frac{\sqrt{3}}{4} a^2$$
Where a$a$ is the side length.
### Core Logic
Let the side of the equilateral triangle be a$a$.
Let theta$\theta$ be the angle between the side AC$AC$ and the parallel line L_2$L_2$.
Then, the angle between the side AB$AB$ and the parallel line L_1$L_1$ can be expressed via alternate geometry. Given distances from A$A$ to the lines form right-angled triangles.
### Step 1: Set up geometric projections
Equilateral triangle between parallel lines diagram for Q4 - JEE Main 2026 Morning
From vertex C$C$ to line passing through A$A$ parallel to L_1, L_2$L_1, L_2$, the perpendicular distance is 3$3$.
In the right triangle formed, we have:
sin theta = frac3a$$\sin \theta = \frac{3}{a}$$
Similarly, point B$B$ lies on L_1$L_1$. The perpendicular distance from A$A$ to L_1$L_1$ is 6$6$.
However, combining the overall heights between the parallel lines, the total distance between L_1$L_1$ and L_2$L_2$ is 6 + 3 = 9$6 + 3 = 9$.
The projection of side BC$BC$ (which connects L_1$L_1$ and L_2$L_2$) vertically is 9$9$.
sin(60^circ + theta) = frac9a$$\sin(60^{\circ} + \theta) = \frac{9}{a}$$
### Step 2: Solve the trigonometric system
Expand sin(60^circ + theta)$\sin(60^{\circ} + \theta)$:
fracsqrt32 cos theta + frac12 sin theta = frac9a$$\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta = \frac{9}{a}$$
Substitute sin theta = frac3a$\sin \theta = \frac{3}{a}$ and cos theta = sqrt1 - sin^2 theta = sqrt1 - frac9a^2$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{9}{a^2}}$:
fracsqrt32 sqrt1 - frac9a^2 + frac12 left(frac3aright) = frac9a$$\frac{\sqrt{3}}{2} \sqrt{1 - \frac{9}{a^2}} + \frac{1}{2} \left(\frac{3}{a}\right) = \frac{9}{a}$$sqrt3 sqrt1 - frac9a^2 + frac3a = frac18a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} + \frac{3}{a} = \frac{18}{a}$$sqrt3 sqrt1 - frac9a^2 = frac15a$$\sqrt{3} \sqrt{1 - \frac{9}{a^2}} = \frac{15}{a}$$
Squaring both sides:
3 left(1 - frac9a^2right) = frac225a^2$$3 \left(1 - \frac{9}{a^2}\right) = \frac{225}{a^2}$$3 - frac27a^2 = frac225a^2$$3 - \frac{27}{a^2} = \frac{225}{a^2}$$3 = frac252a^2 Rightarrow a^2 = 84$$3 = \frac{252}{a^2} \Rightarrow a^2 = 84$$
### Step 3: Calculate Area
textArea of Delta ABC = fracsqrt34 a^2$$\text{Area of } \Delta ABC = \frac{\sqrt{3}}{4} a^2$$= fracsqrt34 times 84 = 21sqrt3$$= \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$$
### Pattern Recognition
When a rigid polygon (like an equilateral triangle or square) is wedged between parallel lines, set a base orientation angle theta$\theta$ for one edge and use rotational shifts (e.g., 60^circ + theta$60^{\circ} + \theta$) to project heights. Expanding the sine addition formula instantly yields the side length.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Class 11 Maths: Trigonometric Functions
Q69jee_main_2025_02_april_eveningEquation of a Straight Line
Let the area of the triangle formed by a straight Line L: x + by + c = 0$L: x + by + c = 0$ with co-ordinate axes be 48 square units. If the perpendicular drawn from the origin to the line L$L$ makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, then the value of b^2 + c^2$b^2 + c^2$ is:
A. 90
B. 93
C. 97
D. 83
Solution
### Related Formula
textNormal form of a straight line: x cos alpha + y sin alpha = p$$\text{Normal form of a straight line: } x \cos \alpha + y \sin \alpha = p$$textArea of right-angled triangle formed with axes: A = frac12 left| x_textintercept cdot y_textintercept right|$$\text{Area of right-angled triangle formed with axes: } A = \frac{1}{2} \left| x_{\text{intercept}} \cdot y_{\text{intercept}} \right|$$
### Core Logic
We write down the normal equation of the straight line using the given polar normal angle alpha = 45^circ$\alpha = 45^\circ$, find its intercept coordinates, and use the area constraint to solve for the coefficients.
### Step 1: Write down normal form
The perpendicular drawn from the origin makes an angle of 45^circ$45^\circ$ with the positive x$x$-axis, so alpha = 45^circ$\alpha = 45^\circ$. The line equation is:
x cos 45^circ + y sin 45^circ = p implies fracxsqrt2 + fracysqrt2 = p$$x \cos 45^\circ + y \sin 45^\circ = p \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = p$$x + y = psqrt2 implies x + y - psqrt2 = 0$$x + y = p\sqrt{2} \implies x + y - p\sqrt{2} = 0$$
Comparing this with the given format x + by + c = 0$x + by + c = 0$, we find:
b = 1 quad textand quad c = -psqrt2$$b = 1 \quad \text{and} \quad c = -p\sqrt{2}$$
### Step 2: Solve for the parameters using the area constraint
The line equation is x + y = psqrt2$x + y = p\sqrt{2}$. The intercepts are:
- x_textintercept = psqrt2$x_{\text{intercept}} = p\sqrt{2}$
- y_textintercept = psqrt2$y_{\text{intercept}} = p\sqrt{2}$
The area of the right-angled triangle formed with the axes is:
textArea = frac12 left| psqrt2 cdot psqrt2 right| = p^2$$\text{Area} = \frac{1}{2} \left| p\sqrt{2} \cdot p\sqrt{2} \right| = p^2$$
Since the area is given as 48 square units:
p^2 = 48$p^2 = 48$
### Step 3: Calculate the requested value
We have:
b^2 = 1^2 = 1$b^2 = 1^2 = 1$c^2 = (-psqrt2)^2 = 2p^2 = 2(48) = 96$$c^2 = (-p\sqrt{2})^2 = 2p^2 = 2(48) = 96$$
Therefore, we find:
b^2 + c^2 = 1 + 96 = 97$$b^2 + c^2 = 1 + 96 = 97$$
### Pattern Recognition
Normal equation coupling: Normal equations of lines xcosalpha+ysinalpha = p$x\cos\alpha+y\sin\alpha = p$ are extremely powerful when normal angles are specified. For alpha=45^circ$\alpha=45^\circ$, the coordinate intercepts are identical, making the area relation A=p^2$A=p^2$ exceptionally simple.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q55jee_main_2025_03_april_eveningFamily of Lines
Consider the lines x(3lambda + 1) + y(7lambda + 2) = 17lambda + 5$x(3\lambda + 1) + y(7\lambda + 2) = 17\lambda + 5$, lambda$\lambda$ being a parameter, all passing through a point P$P$. One of these lines (say L$L$) is farthest from the origin. If the distance of L$L$ from the point (3, 6)$(3, 6)$ is d$d$, then the value of d^2$d^2$ is
A.20$20$
B.30$30$
C.10$10$
D.15$15$
Solution
### Related Formula
A family of lines passing through the intersection of L_1 = 0$L_1 = 0$ and L_2 = 0$L_2 = 0$ is expressed as:
L_1 + lambda L_2 = 0$$L_1 + \lambda L_2 = 0$$
For a point P$P$ through which a family of lines passes, the line in the family that is at the maximum distance from origin O$O$ is the line perpendicular to OP$OP$ passing through P$P$.
### Core Logic
Rearranging the equation of the given lines in terms of lambda$\lambda$:
(x + 2y - 5) + lambda(3x + 7y - 17) = 0$$(x + 2y - 5) + \lambda(3x + 7y - 17) = 0$$
To find point P$P$, solve the system:
1. x + 2y = 5 implies x = 5 - 2y$x + 2y = 5 \implies x = 5 - 2y$
2. 3x + 7y = 17$3x + 7y = 17$
### Step 1: Finding P$P$ and Line L$L$
Substitute x = 5-2y$x = 5-2y$ into the second equation:
3(5 - 2y) + 7y = 17 implies 15 + y = 17 implies y = 2$$3(5 - 2y) + 7y = 17 \implies 15 + y = 17 \implies y = 2$$x = 5 - 2(2) = 1$$x = 5 - 2(2) = 1$$
Thus, the common intersection point is P(1, 2)$P(1, 2)$.
The line farthest from the origin is perpendicular to the segment OP$OP$ joining the origin O(0,0)$O(0,0)$ to P(1,2)$P(1,2)$.
- Slope of OP = frac2 - 01 - 0 = 2$OP = \frac{2 - 0}{1 - 0} = 2$
- Slope of L$L$ (m$m$) = -frac12$= -\frac{1}{2}$
Equation of line L$L$ passing through P(1,2)$P(1,2)$:
y - 2 = -frac12(x - 1) implies 2y - 4 = -x + 1 implies x + 2y - 5 = 0$$y - 2 = -\frac{1}{2}(x - 1) \implies 2y - 4 = -x + 1 \implies x + 2y - 5 = 0$$
### Step 2: Distance calculation from (3,6)$(3,6)$
The distance d$d$ of point Q(3,6)$Q(3,6)$ from line x + 2y - 5 = 0$x + 2y - 5 = 0$ is:
d = left| frac3 + 2(6) - 5sqrt1^2 + 2^2 right| = left| frac10sqrt5 right| = 2sqrt5$$d = \left| \frac{3 + 2(6) - 5}{\sqrt{1^2 + 2^2}} \right| = \left| \frac{10}{\sqrt{5}} \right| = 2\sqrt{5}$$
Calculating d^2$d^2$:
d^2 = (2sqrt5)^2 = 20$$d^2 = (2\sqrt{5})^2 = 20$$
### Pattern Recognition
Short Shortcut: The maximum distance of a family of lines passing through P$P$ from the origin is simply the length OP$OP$. The line perpendicular to OP$OP$ at P$P$ is the unique farthest line.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q70jee_main_2025_07_april_morningOrthocentre of a Triangle
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
A.4$4$
B.10$10$
C.8$8$
D.6$6$
Solution
### Related Formula
The orthocentre P$P$ of a triangle is the point of intersection of its altitudes.
Area of a triangle with a horizontal base lying on the x-axis is:
textArea = frac12 times textbase times textheight = frac12 times |x_C - x_B| times |y_P|$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times |x_C - x_B| \times |y_P|$$
### Core Logic
Find vertex A$A$ by solving the line equations AB$AB$ and AC$AC$:
3y - x = 2 implies x = 3y - 2$$3y - x = 2 \implies x = 3y - 2$$
Substitute into x + y = 2 implies (3y - 2) + y = 2 implies 4y = 4 implies y = 1$x + y = 2 \implies (3y - 2) + y = 2 \implies 4y = 4 \implies y = 1$.
Then x = 3(1) - 2 = 1$x = 3(1) - 2 = 1$. So vertex A$A$ is (1, 1)$(1, 1)$.
Find vertices B$B$ and C$C$ where the lines cross the x-axis (y = 0$y = 0$):
- For B$B$ (on line AB$AB$): 3(0) - x = 2 implies x = -2 implies B(-2, 0)$3(0) - x = 2 \implies x = -2 \implies B(-2, 0)$
- For C$C$ (on line AC$AC$): x + 0 = 2 implies x = 2 implies C(2, 0)$x + 0 = 2 \implies x = 2 \implies C(2, 0)$
Base length BC = |2 - (-2)| = 4$BC = |2 - (-2)| = 4$.
### Step 1: Find Equations of Altitudes
Orthocentre of a Triangle diagram for Q70 - JEE Main 2025 Morning
1. **Altitude from A to BC**:
Since BC$BC$ lies along the x-axis, the altitude from A(1, 1)$A(1, 1)$ must be a vertical line:
textEquation of Altitude 1: x = 1$$\text{Equation of Altitude 1}: x = 1$$
2. **Altitude from B to AC**:
Slope of line AC$AC$ (x + y = 2$x + y = 2$) is m_AC = -1$m_{AC} = -1$.
Therefore, the slope of the altitude perpendicular to AC$AC$ is m_2 = -frac1-1 = 1$m_2 = -\frac{1}{-1} = 1$.
Passing through B(-2, 0)$B(-2, 0)$:
y - 0 = 1(x - (-2)) implies y = x + 2 implies x - y + 2 = 0$$y - 0 = 1(x - (-2)) \implies y = x + 2 \implies x - y + 2 = 0$$
### Step 2: Solve for Orthocentre coordinates P
Intersect the altitude equations: x = 1$x = 1$ and y = x + 2$y = x + 2$:
y = 1 + 2 = 3$y = 1 + 2 = 3$
Hence, the orthocentre is P(1, 3)$P(1, 3)$.
### Step 3: Compute Area of Triangle PBC
Triangle PBC$PBC$ has base BC = 4$BC = 4$ on the x-axis, and vertex P(1, 3)$P(1, 3)$ gives a height of 3$3$.
textArea = frac12 times 4 times 3 = 6$$\text{Area} = \frac{1}{2} \times 4 \times 3 = 6$$
### Pattern Recognition
When a triangle has its base sitting entirely on the coordinate axis, the altitude dropped from the opposite vertex is simplified to a pure horizontal or vertical line path, heavily reducing your linear derivation equation workload.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Qjee_main_2025_08_april_eveningRotation of Coordinate Axes / Distance from Origin
A line passing through the point P(a, 0)$P(a, 0)$ makes an acute angle alpha$\alpha$ with the positive x-axis. Let this line be rotated about the point mathrmP$\mathrm{P}$ through an angle fracalpha2$\frac{\alpha}{2}$ in the clock-wise direction. If in the new position, the slope of the line is 2 - sqrt3$2 - \sqrt{3}$ and its distance from the origin is frac1sqrt2$\frac{1}{\sqrt{2}}$, then the value of 3mathrma^2tan^2alpha - 2sqrt3$3\mathrm{a}^2\tan^2\alpha - 2\sqrt{3}$ is
A.4$4$
B.6$6$
C.5$5$
D.8$8$
Solution
### Related Formula
textDistance from origin d = frac|C|sqrtA^2+B^2$$\text{Distance from origin } d = \frac{|C|}{\sqrt{A^2+B^2}}$$
### Core Logic
Decode slope angle fields to define vector shifts. Match variables to isolate focal root segments accurately as provided in the reference layout steps.
### Step 1: Identify Phase Shift Angles
Given final tracking position slope component is:
m = 2 - sqrt3 = tan 15^circ$$m = 2 - \sqrt{3} = \tan 15^\circ$$
Because clockwise translation turns angular coordinates down by half increments:
alpha - fracalpha2 = 15^circ implies alpha = 30^circ$$\alpha - \frac{\alpha}{2} = 15^\circ \implies \alpha = 30^\circ$$
### Step 2: Construct Rotated Vector Path Line
The line equation through point (a,0)$(a,0)$ with final angle scaling is:
y = (2-sqrt3)(x-a) implies (2-sqrt3)x - y - a(2-sqrt3) = 0$$y = (2-\sqrt{3})(x-a) \implies (2-\sqrt{3})x - y - a(2-\sqrt{3}) = 0$$
### Step 3: Solve Origin Geometric Profiles
Impose perpendicular bounds tracking distance rule:
left| fracsqrt3a - 2asqrt4+3-4sqrt3+1 right| = frac1sqrt2 implies a^2 = 2(2+sqrt3)$$\left| \frac{\sqrt{3}a - 2a}{\sqrt{4+3-4\sqrt{3}+1}} \right| = \frac{1}{\sqrt{2}} \implies a^2 = 2(2+\sqrt{3})$$
Target formulation output text string solution:
3a^2 tan^2 30^circ - 2sqrt3 = 3 times 2(2+sqrt3) times frac13 - 2sqrt3 = 4$$3a^2 \tan^2 30^\circ - 2\sqrt{3} = 3 \times 2(2+\sqrt{3}) \times \frac{1}{3} - 2\sqrt{3} = 4$$
{{SOL_IMG_62}}
### Pattern Recognition
Recognizing standard trigonometric slopes (tan 15^circ = 2-sqrt3$\tan 15^\circ = 2-\sqrt{3}$) bypasses computational bottlenecks instantly when working with line updates.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
More Straight Lines Questions — jee_main_2024_29_jan_morning
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