In a Δ 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 (α,β)$(\alpha,\beta)$, then α+2β$\alpha+2\beta$ is equal to
A.42$42$
B.39$39$
C.48$48$
D.45$45$
Solution & Explanation
Related Formula
Image of a point (x₁, y₁)$(x_1, y_1)$ across line y=x$y=x$ is (y₁, x₁)$(y_1, x_1)$.
Angle Bisector Theorem: The angle bisector of a triangle divides the opposite side into segments proportional to the lengths of the adjacent sides:
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 ⇒ x = 2 ⇒ y = 2$$2x - x = 2 \Rightarrow x = 2 \Rightarrow y = 2$$
Given 2AB = BC$2AB = BC$, so (AB)/(BC) = (1)/(2)$\frac{AB}{BC} = \frac{1}{2}$.
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 (xc, yc)$(x_c, y_c)$.
Applying the section formula for D(2,2)$D(2,2)$ dividing A(4,6)$A(4,6)$ and C(xc, yc)$C(x_c, y_c)$ in ratio 1:2$1:2$:
Vertex B(α, β)$B(\alpha, \beta)$ 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$5x - 4y - 14 = 0$:
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.
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 — Page 4
Q65jee_main_2025_04_april_morningOrthocentre of a Triangle
Let the three sides of a triangle be on the lines 4x - 7y + 10 = 0$4x - 7y + 10 = 0$, x + y = 5$x + y = 5$ and 7x + 4y = 15$7x + 4y = 15$. Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x = 0$x = 0$, y = 0$y = 0$ and x + y = 1$x + y = 1$ is
A. 5
B.√(5)$\sqrt{5}$
C.√(20)$\sqrt{20}$
D. 20
Solution
Related Formula
For any right-angled triangle, the orthocentre lies exactly at the vertex containing the 90°$90^{\circ}$ right angle.
Distance formula:
Thus, Triangle 1 is a right-angled triangle. Its orthocentre B$B$ is the intersection point of L₁$L_1$ and L₂$L_2$:
Solving 4x - 7y = -10$4x - 7y = -10$ and 7x + 4y = 15$7x + 4y = 15$:
Multiplying first by 4, second by 7, and adding yields x = 1, y = 2 B(1, 2)$x = 1, y = 2 \implies B(1, 2)$.
Orthocentre of a Triangle diagram for Q65 - JEE Main 2025 Morning
Step 1: Locate Second Orthocentre
Triangle 2 is formed by x = 0$x = 0$, y = 0$y = 0$, and x + y = 1$x + y = 1$. This is a right triangle with vertices at (0,0), (1,0), (0,1)$(0,0), (1,0), (0,1)$.
The right-angled vertex is at the origin P(0, 0)$P(0, 0)$. Thus, its orthocentre is P(0, 0)$P(0, 0)$.
Orthocentre of a Triangle diagram for Q65 - JEE Main 2025 Morning
Step 2: Distance Computation
Find the distance between B(1,2)$B(1,2)$ and P(0,0)$P(0,0)$:
Always check for mutually perpendicular side orientations (m₁ · m₂ = -1$m_1 \cdot m_2 = -1$) when finding orthocentres in competitive math papers. This completely cuts out lengthy altitude equation derivation tracks.
Chapter Mix
Class 11 Mathematics: Straight Lines
Q51jee_main_2025_07_april_eveningOrthocentre of a Triangle
If the orthocentre of the triangle formed by the lines y = x + 1$y = x + 1$, y = 4x - 8$y = 4x - 8$ and y = mx + c$y = mx + c$ is at (3, -1)$(3, -1)$, then m - c$m - c$ is:
A.0$0$
B.-2$-2$
C.4$4$
D.2$2$
Solution
Related Formula
The product of slopes of two mutually perpendicular lines is always equal to -1$-1$:
m₁ · m₂ = -1$$m_1 \cdot m_2 = -1$$
Core Logic
Let the vertices of the triangle be P$P$, Q$Q$, and R$R$. The lines given are:
y = x + 1$y = x + 1$
y = 4x - 8$y = 4x - 8$
y = mx + c$y = mx + c$
Solving lines y = x + 1$y = x + 1$ and y = 4x - 8$y = 4x - 8$ gives the vertex P(3, 4)$P(3, 4)$.
The orthocentre is given as H(3, -1)$H(3, -1)$. Notice that the x$x$-coordinate of P$P$ and H$H$ are identical (x = 3$x = 3$). This implies that the altitude from vertex P$P$ to the base line y = mx + c$y = mx + c$ is a vertical line along x = 3$x = 3$.
Orthocentre of a Triangle diagram for Q51 - JEE Main 2025 Evening
Step 1: Determine the Slopes
Since the altitude from P$P$ is vertical, the side opposite to it (which lies on y = mx + c$y = mx + c$) must be a horizontal line.
Therefore, the slope of the line y = mx + c$y = mx + c$ must be zero:
m = 0$m = 0$
Step 2: Solve for c
Let's find point Q$Q$ by intersecting y = x + 1$y = x + 1$ and y = mx + c$y = mx + c$. Since m = 0$m = 0$, y = c$y = c$, we get Q(c-1, c)$Q(c-1, c)$.
Using the property that the line segment connecting Q$Q$ to the opposite side's altitude is perpendicular to line PR$PR$ (y = 4x - 8$y = 4x - 8$):
Slope of QH · Slope of PR = -1$$\text{Slope of } QH \cdot \text{Slope of } PR = -1$$(-1 - c)/(3 - (c - 1)) · 4 = -1$$\frac{-1 - c}{3 - (c - 1)} \cdot 4 = -1$$(-4(c + 1))/(4 - c) = -1 4c + 4 = 4 - c 5c = 0 c = 0$$\frac{-4(c + 1)}{4 - c} = -1 \implies 4c + 4 = 4 - c \implies 5c = 0 \implies c = 0$$
Step 3: Evaluate m - c
Substituting the values of m$m$ and c$c$:
m - c = 0 - 0 = 0$$m - c = 0 - 0 = 0$$
Pattern Recognition
When the x$x$-coordinate of a vertex matches the x$x$-coordinate of the orthocentre, the altitude is vertical, forcing the opposite base to be purely horizontal (m=0$m=0$). This observation cuts down calculation time completely.
Chapter Mix
Class 11 Mathematics: Straight Lines
Qjee_main_2025_24_jan_morningConcurrency of Straight Lines
Let the lines 3x - 4y - α = 0$3x - 4y - \alpha = 0$, 8x - 11y - 33 = 0$8x - 11y - 33 = 0$, and 2x - 3y + λ = 0$2x - 3y + \lambda = 0$ be concurrent. If the image of the point (1, 2)$(1, 2)$ in the line 2x - 3y + λ = 0$2x - 3y + \lambda = 0$ is ((57)/(13),(-40)/(13))$\left(\frac{57}{13},\frac{-40}{13}\right)$ , then |α λ|$|\alpha \lambda|$ is equal to :
A.84$84$
B.91$91$
C.113$113$
D.101$101$
Solution
Related Formula
The midpoint between a point and its reflection image must lie exactly on the line mirror equation.
Core Logic
Find the midpoint M$M$ between point P(1, 2)$P(1, 2)$ and its given reflection image Q((57)/(13), (-40)/(13))$Q\left(\frac{57}{13}, \frac{-40}{13}\right)$:
Using the midpoint property to evaluate unknown line parameters from reflection images is often much faster than using full distance formulas.
Chapter Mix
Class 11 Mathematics: Straight Lines
Q67jee_main_2025_28_jan_eveningAngle Between Lines
Two equal sides of an isosceles triangle are along -x+2y=4$-x+2y=4$ and x+y=4.$x+y=4.$ If m is the slope of its third side, then the sum, of all possible distinct values of m, is:
A.-6$-6$
B.12$12$
C.6$6$
D.-2√(10)$-2\sqrt{10}$
Solution
Related Formula
Angle θ$\theta$ between two lines with slopes m₁$m_1$ and m₂$m_2$:
The quadratic equation for m$m$ is m² - 6m - 1 = 0$m^2 - 6m - 1 = 0$.
The sum of possible distinct values of m$m$ is given by the sum of roots of this quadratic:
Sum of roots = -(-6)/(1) = 6$$\text{Sum of roots} = -\frac{-6}{1} = 6$$
Pattern Recognition
Instead of solving for the explicit values of the slopes (which involve radicals), using Vieta's relations directly on the quadratic equation m² - 6m - 1 = 0$m^2 - 6m - 1 = 0$ gives the final answer instantly.
Chapter Mix
Class 11 Mathematics: Straight Lines
Q54jee_main_2025_29_jan_morningCentroid and Image of a Point
Let ABC be a triangle formed by the lines 7x - 6y + 3 = 0, x + 2y - 31 = 0$7\mathrm{x} - 6\mathrm{y} + 3 = 0, \mathrm{x} + 2\mathrm{y} - 31 = 0$ and 9x - 2y - 19 = 0$9\mathrm{x} - 2\mathrm{y} - 19 = 0$ . Let the point (h,k)$(\mathrm{h},\mathrm{k})$ be the image of the centroid of Δ ABC$\Delta ABC$ in the line 3x + 6y - 53 = 0$3\mathrm{x} + 6\mathrm{y} - 53 = 0$ . Then \mathrm{h}^2 + \mathrm{k}^2 + \mathrm{hk} is equal to
A. 37
B. 47
C. 40
D. 36
Solution
Related Formula
Centroid G = ((x₁+x₂+x₃)/(3), (y₁+y₂+y₃)/(3))$$\text{Centroid } G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)$$Image of (x₁, y₁) in line ax+by+c=0: (x-x₁)/(a) = (y-y₁)/(b) = -2(ax₁+by₁+c)/(a²+b²)$$\text{Image of } (x_1, y_1) \text{ in line } ax+by+c=0: \frac{x-x_1}{a} = \frac{y-y_1}{b} = -2\frac{ax_1+by_1+c}{a^2+b^2}$$
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
