Let A(1, 2)$A(1, 2)$ and C(-3, -6)$C(-3, -6)$ be two diagonally opposite vertices of a rhombus, whose sides AD$AD$ and BC$BC$ are parallel to the line 7x - y = 14$7x - y = 14$. If B(alpha, beta)$B(\alpha, \beta)$ and D(gamma, delta)$D(\gamma, \delta)$ are the other two vertices, then |alpha + beta + gamma + delta|$|\alpha + \beta + \gamma + \delta|$ is equal to :
A.9$9$
B.3$3$
C.6$6$
D.1$1$
Solution & Explanation
### Related Formula
In a rhombus, diagonals bisect each other. Therefore, the midpoint of diagonal AC$AC$ is the same as the midpoint of diagonal BD$BD$.
### Core Logic
Properties of Rhombus diagram for Q4 - JEE Main 2026 Evening
Given the coordinates of vertices A(1, 2)$A(1, 2)$ and C(-3, -6)$C(-3, -6)$, we can find the midpoint O$O$ of the diagonal AC$AC$.
x_m = frac1 + (-3)2 = -1$$x_m = \frac{1 + (-3)}{2} = -1$$y_m = frac2 + (-6)2 = -2$$y_m = \frac{2 + (-6)}{2} = -2$$
So, the midpoint is (-1, -2)$(-1, -2)$.
Let the coordinates of B$B$ and D$D$ be (alpha, beta)$(\alpha, \beta)$ and (gamma, delta)$(\gamma, \delta)$ respectively. Since the midpoint of BD$BD$ is also (-1, -2)$(-1, -2)$:
fracalpha+gamma2 = -1 implies alpha+gamma = -2$$\frac{\alpha+\gamma}{2} = -1 \implies \alpha+\gamma = -2$$fracbeta+delta2 = -2 implies beta+delta = -4$$\frac{\beta+\delta}{2} = -2 \implies \beta+\delta = -4$$
### Step 1: Final Calculation
We need the absolute value of the sum of the coordinates of B$B$ and D$D$:
|alpha + beta + gamma + delta| = |(alpha + gamma) + (beta + delta)|$$|\alpha + \beta + \gamma + \delta| = |(\alpha + \gamma) + (\beta + \delta)|$$= |-2 - 4| = |-6| = 6$$= |-2 - 4| = |-6| = 6$$
### Pattern Recognition
The extra information regarding the parallel line 7x - y = 14$7x - y = 14$ is completely redundant for finding the sum of the coordinates. Always check if a basic geometric property (like diagonals bisecting) circumvents heavy calculations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Keywords:#diagonally opposite vertices of a rhombus#JEE Main 2026 Evening Q4#Straight Lines JEE Main 2026#Properties of Rhombus JEE Main 2026
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
Q7jee_main_2026_22_january_eveningOrthocenter and Concurrency
Among the statements:
(S1): If A(5,-1)$A(5,-1)$ and B(-2,3)$B(-2,3)$ are two vertices of a triangle, whose orthocentre is (0,0)$(0,0)$, then its third vertex is (-4,-7)$(-4,-7)$ and
(S2): If positive numbers 2a, b, c$2a, b, c$ are three consecutive terms of an A.P., then the lines ax + by + c = 0$ax + by + c = 0$ are concurrent at (2,-2)$(2,-2)$.
A. Only (S1) is correct
B. Only (S2) is correct
C. Both are incorrect
D. Both are correct
Solution
### Related Formula
Orthocenter property: Altitude from A$A$ is perpendicular to BC$BC$, Altitude from B$B$ is perpendicular to AC$AC$.
Concurrency of lines: ax+by+c=0$ax+by+c=0$ passes through (x_0, y_0)$(x_0, y_0)$ if ax_0+by_0+c=0$ax_0+by_0+c=0$ holds.
### Core Logic
Triangle orthocenter diagram for Q7 - JEE Main 2026 Evening
Statement 1:
Let third vertex be C(h,k)$C(h,k)$ and orthocenter O(0,0)$O(0,0)$.
- Slope AO cdot$AO \cdot$ Slope BC = -1 implies left(frac-15right) cdot left(frack-3h+2right) = -1 implies 5h - k + 13 = 0$BC = -1 \implies \left(\frac{-1}{5}\right) \cdot \left(\frac{k-3}{h+2}\right) = -1 \implies 5h - k + 13 = 0$
- Slope BO cdot$BO \cdot$ Slope AC = -1 implies left(frac3-2right) cdot left(frack+1h-5right) = -1 implies 4k = 7h$AC = -1 \implies \left(\frac{3}{-2}\right) \cdot \left(\frac{k+1}{h-5}\right) = -1 \implies 4k = 7h$
Solving simultaneously gives h = -4, k = -7 implies C(-4,-7)$h = -4, k = -7 \implies C(-4,-7)$. Statement 1 is correct.
Statement 2:
2a, b, c$2a, b, c$ in A.P. implies 2b = 2a + c implies 2a - 2b + c = 0$\implies 2b = 2a + c \implies 2a - 2b + c = 0$.
Comparing with ax + by + c = 0$ax + by + c = 0$ gives x = 2, y = -2$x = 2, y = -2$. Statement 2 is correct.
### Step 1: Final Conclusion
Both (S1) and (S2) are correct statements.
### Pattern Recognition
Orthocenter coordinates yield perpendicularity conditions via slopes. Concurrency follows directly from linear relation between line coefficients.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Q4jee_main_2026_23_january_morningArea and Distance
A rectangle is formed by the lines x = 0, y = 0, x = 3$x = 0, y = 0, x = 3$ and y = 4$y = 4$. Let the line L$L$ be perpendicular to 3x + y + 6 = 0$3x + y + 6 = 0$ and divide the area of the rectangle into two equal parts. Then the distance of the point left(frac12, -5right)$\left(\frac{1}{2}, -5\right)$ from the line L$L$ is equal to:
A.2sqrt5$2\sqrt{5}$
B.3sqrt10$3\sqrt{10}$
C.sqrt10$\sqrt{10}$
D.2sqrt10$2\sqrt{10}$
Solution
### Related Formula
textDistance = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$$\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$
### Core Logic
Any line that divides the area of a rectangle into two equal parts must pass through the center of the rectangle.
Area and Distance diagram for Q4 - JEE Main 2026 Morning
The vertices of the rectangle are (0,0)$(0,0)$, (3,0)$(3,0)$, (3,4)$(3,4)$, and (0,4)$(0,4)$. Its center is at left(frac32, 2right)$\left(\frac{3}{2}, 2\right)$.
### Step 1: Equation of Line L
The line L$L$ is perpendicular to 3x + y + 6 = 0$3x + y + 6 = 0$. The slope of this reference line is -3$-3$. Thus, the slope of L$L$ is m = frac13$m = \frac{1}{3}$.
The equation of L$L$ passing through left(frac32, 2right)$\left(\frac{3}{2}, 2\right)$ is:
y - 2 = frac13left(x - frac32right)$$y - 2 = \frac{1}{3}\left(x - \frac{3}{2}\right)$$y = fracx3 - frac12 + 2 Rightarrow y = fracx3 + frac32$$y = \frac{x}{3} - \frac{1}{2} + 2 \Rightarrow y = \frac{x}{3} + \frac{3}{2}$$
Multiplying by 6 gives:
6y = 2x + 9 Rightarrow 2x - 6y + 9 = 0$$6y = 2x + 9 \Rightarrow 2x - 6y + 9 = 0$$
### Step 2: Calculate Distance
We need the perpendicular distance from left(frac12, -5right)$\left(\frac{1}{2}, -5\right)$ to the line 2x - 6y + 9 = 0$2x - 6y + 9 = 0$:
D = fracleft| 2left(frac12right) - 6(-5) + 9 right|sqrt2^2 + (-6)^2$$D = \frac{\left| 2\left(\frac{1}{2}\right) - 6(-5) + 9 \right|}{\sqrt{2^2 + (-6)^2}}$$D = frac|1 + 30 + 9|sqrt4 + 36 = frac40sqrt40 = sqrt40 = 2sqrt10$$D = \frac{|1 + 30 + 9|}{\sqrt{4 + 36}} = \frac{40}{\sqrt{40}} = \sqrt{40} = 2\sqrt{10}$$
### Pattern Recognition
Area bisectors for symmetric geometric figures (rectangles, circles, ellipses) always pass exactly through their geometric center.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
Q13jee_main_2026_24_january_morningAngle Bisectors of a Triangle
Let A(1, 0)$A(1, 0)$, B(2, -1)$B(2, -1)$ and Cleft(frac73, frac43right)$C\left(\frac{7}{3}, \frac{4}{3}\right)$ be three points. If the equation of the bisector of the angle ABC$ABC$ is alpha x + beta y = 5$\alpha x + \beta y = 5$, then the value of alpha^2 + beta^2$\alpha^2 + \beta^2$ is
A.8$8$
B.5$5$
C.13$13$
D.10$10$
Solution
### Related Formula
Angle bisector divides the opposite side in the ratio of adjacent sides: fracBDDC = fracABBC$\frac{BD}{DC} = \frac{AB}{BC}$ (Internal Bisector Theorem).
### Core Logic
Angle Bisector diagram for Q13
Calculate lengths AB$AB$ and BC$BC$ (wait, we need bisector of angle ABC$ABC$, so it intersects AC$AC$ at D$D$, so fracADDC = fracABBC$\frac{AD}{DC} = \frac{AB}{BC}$).
### Step 1: Side Lengths
AB = sqrt(2-1)^2 + (-1-0)^2 = sqrt1 + 1 = sqrt2$$AB = \sqrt{(2-1)^2 + (-1-0)^2} = \sqrt{1 + 1} = \sqrt{2}$$BC = sqrtleft(frac73-2right)^2 + left(frac43-(-1)right)^2 = sqrtfrac19 + frac499 = sqrtfrac509 = frac5sqrt23$$BC = \sqrt{\left(\frac{7}{3}-2\right)^2 + \left(\frac{4}{3}-(-1)\right)^2} = \sqrt{\frac{1}{9} + \frac{49}{9}} = \sqrt{\frac{50}{9}} = \frac{5\sqrt{2}}{3}$$
Ratio fracABBC = fracsqrt2frac5sqrt23 = frac35$\frac{AB}{BC} = \frac{\sqrt{2}}{\frac{5\sqrt{2}}{3}} = \frac{3}{5}$.
### Step 2: Coordinates of D
Point D$D$ divides AC$AC$ internally in ratio 3:5$3:5$.
x = frac3(7/3) + 5(1)3+5 = frac7+58 = frac128 = frac32$$x = \frac{3(7/3) + 5(1)}{3+5} = \frac{7+5}{8} = \frac{12}{8} = \frac{3}{2}$$y = frac3(4/3) + 5(0)3+5 = frac48 = frac12$$y = \frac{3(4/3) + 5(0)}{3+5} = \frac{4}{8} = \frac{1}{2}$$D = left(frac32, frac12right)$D = \left(\frac{3}{2}, \frac{1}{2}\right)$
### Step 3: Equation of Bisector
The bisector passes through B(2,-1)$B(2,-1)$ and D(3/2, 1/2)$D(3/2, 1/2)$.
Slope m = frac1/2 - (-1)3/2 - 2 = frac3/2-1/2 = -3$m = \frac{1/2 - (-1)}{3/2 - 2} = \frac{3/2}{-1/2} = -3$
Equation: y - (-1) = -3(x - 2)$y - (-1) = -3(x - 2)$y + 1 = -3x + 6 Rightarrow 3x + y = 5$y + 1 = -3x + 6 \Rightarrow 3x + y = 5$Angle Bisector diagram for Q13
### Step 4: Final Value Calculation
Comparing 3x + y = 5$3x + y = 5$ with alpha x + beta y = 5$\alpha x + \beta y = 5$, we get alpha = 3$\alpha = 3$, beta = 1$\beta = 1$.
alpha^2 + beta^2 = 9 + 1 = 10$$\alpha^2 + \beta^2 = 9 + 1 = 10$$
### Pattern Recognition
When asked for a specific angle bisector in a coordinate triangle, using the Angle Bisector Theorem (ratio trick) to find a second point is much faster than computing formulas of bisectors between lines.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Straight Lines
More Straight Lines Questions — jee_main_2026_23_january_evening
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