Let ABC$ABC$ be an equilateral triangle with orthocenter at the origin and the side BC$BC$ on the line x + 2√(2)y = 4$x + 2\sqrt{2}y = 4$. If the co-ordinates of the vertex A$A$ are (α, β)$(\alpha, \beta)$, then the greatest integer less than or equal to |α + √(2)β|$|\alpha + \sqrt{2}\beta|$ is
A.2$2$
B.3$3$
C.5$5$
D.4$4$
Solution & Explanation
Core Logic
Properties of Triangles
For an equilateral triangle, the orthocenter coincides with the centroid O(0,0)$O(0,0)$.
Let AD$AD$ be the altitude from A$A$ to side BC$BC$. The line AD$AD$ is perpendicular to BC$BC$.
Equation of BC$BC$: x + 2√(2)y - 4 = 0$x + 2\sqrt{2}y - 4 = 0$
Slope of BC$BC$: mBC = - 12√(2)$m_{BC} = -\frac{1}{2\sqrt{2}}$
Since AD ⊥ BC$AD \perp BC$, mBC · mAD = -1$m_{BC} \cdot m_{AD} = -1$
Since A$A$ and the origin O$O$ must lie on opposite sides of BC$BC$ (wait, O$O$ is inside the triangle, so A$A$ and O$O$ lie on opposite sides of BC$BC$? No, O$O$ is inside the triangle, so the origin and vertex A are on opposite sides of the chord BC$BC$ if we look from the circumcenter. Wait, O(0,0)$O(0,0)$ gives 0+0-4 = -4 < 0$0+0-4 = -4 < 0$. If α = 16/9, β = 32√(2)/9$\alpha = 16/9, \beta = 32\sqrt{2}/9$, then 16/9 + 2√(2)(32√(2)/9) - 4 = 16/9 + 128/9 - 36/9 = 108/9 = 12 > 0$16/9 + 2\sqrt{2}(32\sqrt{2}/9) - 4 = 16/9 + 128/9 - 36/9 = 108/9 = 12 > 0$. Thus, they lie on opposite sides, which is correct for altitude line. Wait, our source notes: A(α,β)$A(\alpha,\beta)$ and (0,0)$(0,0)$ lie on the SAME side of the given line is Rejected. Actually, they lie on opposite sides relative to BC$BC$. Thus (α, β) = (-(8)/(9), -16√(2)9)$(\alpha, \beta) = \left(-\frac{8}{9}, \frac{-16\sqrt{2}}{9}\right)$ is correct because O$O$ is the centroid, so moving from D$D$ to O$O$ and then to A$A$ implies O$O$ is between A$A$ and D$D$. Let's trust the solved matrix: A(α, β) = (-(8)/(9), -16√(2)9)$A(\alpha, \beta) = \left(-\frac{8}{9}, \frac{-16\sqrt{2}}{9}\right)$.
Step 3: Final Value Evaluation
We need the greatest integer less than or equal to |α + √(2)β|$|\alpha + \sqrt{2}\beta|$:
Keywords:#orthocenter of equilateral triangle#JEE Main 2026 Morning Q6#Straight Lines JEE Main 2026#Properties of Triangles JEE Main 2026
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.
