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 & Explanation
### Related Formula
The product of slopes of two mutually perpendicular lines is always equal to -1$-1$:
m_1 cdot m_2 = -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:
1) y = x + 1$y = x + 1$
2) y = 4x - 8$y = 4x - 8$
3) 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$y = 4x - 8$):
textSlope of QH cdot textSlope of PR = -1$\text{Slope of } QH \cdot \text{Slope of } PR = -1$frac-1 - c3 - (c - 1) cdot 4 = -1$\frac{-1 - c}{3 - (c - 1)} \cdot 4 = -1$frac-4(c + 1)4 - c = -1 implies 4c + 4 = 4 - c implies 5c = 0 implies 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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Keywords:#orthocentre of the triangle formed by lines#JEE Main 2025 Evening Q51#Straight Lines JEE Main 2025#Orthocentre of a Triangle JEE Main 2025
More Straight Lines Previous-Year Questions — Page 2
Q582025Area of Triangles and Inscribed Shapes
Let the line x + y = 1$x + y = 1$ meet the axes of x$x$ and y$y$ at A and B, respectively. A right angled triangle AMN is inscribed in the triangle OAB, where O is the origin and the points M and N lie on the lines OB and AB, respectively. If the area of the triangle AMN is frac49$\frac{4}{9}$ of the area of the triangle OAB and AN: NB = \lambda : 1, then the sum of all possible value(s) of lambda$\lambda$ is :
A.frac12$\frac{1}{2}$
B.frac136$\frac{13}{6}$
C.frac52$\frac{5}{2}$
D.2$2$
Solution
### Related Formula
Area of a right-angled triangle:
textArea = frac12 times textbase times textheight$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
### Core Logic
The line equation is x + y = 1$x + y = 1$, giving intercept coordinates A(1, 0)$A(1, 0)$ and B(0, 1)$B(0, 1)$.
Area of Triangles and Inscribed Shapes diagram for Q58 - JEE Main 2025 Evening
Area of Delta OAB = frac12 times 1 times 1 = frac12$\Delta OAB = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$.
Given area condition:
textArea of Delta AMN = frac49 times frac12 = frac29$\text{Area of } \Delta AMN = \frac{4}{9} \times \frac{1}{2} = \frac{2}{9}$
### Step 1: Set up Trigonometric Tracing
Let angle MAO = 45^circ - theta$\angle MAO = 45^\circ - \theta$. Since Delta OAB$\Delta OAB$ is isosceles right-angled at O$O$, angle OAB = 45^circ$\angle OAB = 45^\circ$.
This establishes:
OA = 1, quad AM = sec(45^circ - theta)$OA = 1, \quad AM = \sec(45^\circ - \theta)$AN = sec(45^circ - theta)costheta$AN = \sec(45^\circ - \theta)\cos\theta$MN = sec(45^circ - theta)sintheta$MN = \sec(45^\circ - \theta)\sin\theta$
### Step 2: Solve for Angles and Ratios
textArea(Delta AMN) = frac12 times sec^2(45^circ - theta)sinthetacostheta = frac29$\text{Area}(\Delta AMN) = \frac{1}{2} \times \sec^2(45^\circ - \theta)\sin\theta\cos\theta = \frac{2}{9}$
Solving the trigonometric ratio yields:
tantheta = 2 quad textor quad frac12$\tan\theta = 2 \quad \text{or} \quad \frac{1}{2}$
Rejecting tantheta = 2$\tan\theta = 2$ based on physical boundaries within the triangle limits:
fracANNB = fraclambda1 = cottheta = 2$\frac{AN}{NB} = \frac{\lambda}{1} = \cot\theta = 2$
Thus, the valid evaluation matches the option sequence value of 2.
### Pattern Recognition
When dealing with inscribed right triangles inside symmetric linear bounds, parameterizing coordinates with angles matching the axis slope simplifies configuration variables dramatically.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q722025Distance of a Point from a Line
If alpha = 1 + sum_r=1^6 (-3)^r-1 ^12 C_2r-1$\alpha = 1 + \sum_{r=1}^{6} (-3)^{r-1} {}^{12} C_{2r-1}$, then the distance of the point (12, sqrt3)$(12, \sqrt{3})$ from the line alpha x - sqrt3 y + 1 = 0$\alpha x - \sqrt{3} y + 1 = 0$ is
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
Perpendicular distance from a point (x_1, y_1)$(x_1, y_1)$ to a line Ax + By + C = 0$Ax + By + C = 0$:
d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$
### Core Logic
Rewrite the summation for alpha$\alpha$ by embedding complex roots to isolate alternating terms:
alpha = 1 + frac1sqrt3i left[ frac(1 + sqrt3i)^12 - (1 - sqrt3i)^122 right]$\alpha = 1 + \frac{1}{\sqrt{3}i} \left[ \frac{(1 + \sqrt{3}i)^{12} - (1 - \sqrt{3}i)^{12}}{2} \right]$
Using complex cube roots of unity conversions (1 + sqrt3i = -2omega^2$1 + \sqrt{3}i = -2\omega^2$), this simplifies directly to alpha = 1$\alpha = 1$.
### Step 1: Calculating Perpendicular Distance
Substitute alpha = 1$\alpha = 1$ into the line equation, yielding x - sqrt3y + 1 = 0$x - \sqrt{3}y + 1 = 0$.
Evaluating the distance from point (12, sqrt3)$(12, \sqrt{3})$:
d = frac|12 - sqrt3(sqrt3) + 1|sqrt1^2 + (-sqrt3)^2 = frac|12 - 3 + 1|2 = frac102 = 5$d = \frac{|12 - \sqrt{3}(\sqrt{3}) + 1|}{\sqrt{1^2 + (-\sqrt{3})^2}} = \frac{|12 - 3 + 1|}{2} = \frac{10}{2} = 5$
### Pattern Recognition
Binomial series containing terms like sqrt3^2r-1$\sqrt{3}^{2r-1}$ collapse neatly into simple geometric forms using standard complex rotations (e^ipi/3$e^{i\pi/3}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Binomial Theorem
Class 11 Maths: Straight Lines
Q632025Distance between Parallel Lines
A line passes through the origin and makes equal angles with the positive coordinate axes[cite: 602]. It intersects the lines L_1:2x+y+6=0$L_{1}:2x+y+6=0$ and L_2:4x+2y-p=0$L_{2}:4x+2y-p=0$, p>0$p>0$, at the points A$A$ and B$B$, respectively[cite: 603]. If AB = frac9sqrt2$AB = \frac{9}{\sqrt{2}}$ [cite: 605] and the foot of the perpendicular from the point A$A$ on the line L_2$L_{2}$ is M$M$ [cite: 606, 607], then fracAMBM$\frac{AM}{BM}$ is equal to[cite: 619, 620]:
A. 5
B. 4
C. 2
D. 3
Solution
### Related Formula
Slope angle interaction tracking: In right triangle triangle AMB$\triangle AMB$, the tangent of inclination satisfies:
tan theta = fracAMBM$\tan \theta = \frac{AM}{BM}$Distance between Parallel Lines diagram for Q63 - JEE Main 2025 Morning
### Core Logic
A line tracking equal angles through positive coordinate frames has equation y=x$y=x$ [cite: 1353].
Its slope parameter is m_1 = 1$m_1 = 1$ [cite: 1353].
The parallel boundary lines L_1$L_1$ and L_2$L_2$ have a uniform slope value of m_2 = -2$m_2 = -2$ [cite: 1353].
Let theta$\theta$ represent the precise geometric crossing angle matching the line y=x$y=x$ intersecting line L_2$L_2$[cite: 1353].
### Step 1: Finding the requested ratio
Using the slope interaction formula to calculate the tangent angle metric[cite: 1353]:
tan theta = left| fracm_1 - m_21 + m_1 m_2 right|$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ [cite: 1353]
tan theta = left| frac1 - (-2)1 + (1)(-2) right| = left| frac3-1 right| = 3$\tan \theta = \left| \frac{1 - (-2)}{1 + (1)(-2)} \right| = \left| \frac{3}{-1} \right| = 3$ [cite: 1353]
By observing right-angled geometry configuration triangle AMB$\triangle AMB$ [cite: 1353]:
fracAMBM = tan theta = 3$\frac{AM}{BM} = \tan \theta = 3$ [cite: 1353]
### Pattern Recognition
Ratios of side projections of lines intersecting parallel setups depend completely on direction orientations, bypassing raw variable solving steps entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
Q652025Orthocentre 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.sqrt5$\sqrt{5}$
C.sqrt20$\sqrt{20}$
D. 20
Solution
### Related Formula
For any right-angled triangle, the orthocentre lies exactly at the vertex containing the 90^circ$90^{\circ}$ right angle.
Distance formula:
d = sqrt(x_2 - x_1)^2 + (y_2 - y_1)^2$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
### Core Logic
Analyze slopes of lines forming Triangle 1:
L_1: 4x - 7y + 10 = 0 implies m_1 = frac47$L_1: 4x - 7y + 10 = 0 \implies m_1 = \frac{4}{7}$L_2: 7x + 4y - 15 = 0 implies m_2 = -frac74$L_2: 7x + 4y - 15 = 0 \implies m_2 = -\frac{7}{4}$
Notice m_1 cdot m_2 = left(frac47right)left(-frac74right) = -1$m_1 \cdot m_2 = \left(\frac{4}{7}\right)\left(-\frac{7}{4}\right) = -1$.
Thus, Triangle 1 is a right-angled triangle. Its orthocentre B$B$ is the intersection point of L_1$L_1$ and L_2$L_2$:
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 implies 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)$P(0,0)$:
d = sqrt(1 - 0)^2 + (2 - 0)^2 = sqrt1 + 4 = sqrt5$d = \sqrt{(1 - 0)^2 + (2 - 0)^2} = \sqrt{1 + 4} = \sqrt{5}$
### Pattern Recognition
Always check for mutually perpendicular side orientations (m_1 cdot m_2 = -1$m_1 \cdot m_2 = -1$) when finding orthocentres in competitive math papers. This completely cuts out lengthy altitude equation derivation tracks.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Straight Lines
More Straight Lines Questions — jee_main_2025_07_april_evening
We Map Every Repeating Question in Competitive Exams.
Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.
Select Your Target Exam
Choose an exam track below to find formulas per chapter and patterns.
Syncing Exam Intelligence
Mapping formulas and patterns across all tracks…
PATH A — FULL LENGTH PRACTICE
Full Mock Test Hub
Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.
Under Development
PATH B — TARGETED PRACTICE
Topic-wise Practice Hub
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.
