JEE Main 2026 (22 January Shift 2)

LEVELJEE Advanced

Animated Solution for Mathematics - Straight Lines: Among the statements (S1) : If $A (5, - 1)$ and $B (- 2, 3)$ are two vertices of a triangle, whose orthocentre is $(0, 0)$ , then its third vertex is $(- 4, - 7)$ and (S2) : If positive numbers 2a, b, c are three consecutive terms of an A.P., then the lines $a x + b y + c = 0$ are concurrent at $(2, - 2)$ ,

Q: How do I know the lines are concurrent at $(2, -2)$?

The equation $ax + by + c = 0$ represents a family of lines. If you can express this as $a(2) + b(-2) + c = 0$, you have essentially shown that the point $(2, -2)$ satisfies the equation for *any* $a, b, c$ that satisfy the A.P. condition. It is a beautiful way to prove concurrency without finding the intersection of two specific lines.

Select Answer:

Visualized Solution

Introduction to Statement $S 1$

Statement $S 1$ :
Vertices: $A (5, - 1)$ , $B (- 2, 3)$
Orthocenter: $O (0, 0)$
Claim: Third vertex $C$ is $(- 4, - 7)$

The Orthocenter Property

Let the third vertex be $C (h, k)$ .
Property: The line from a vertex to the orthocenter is perpendicular to the opposite side.
$A O ⊥ B C \Rightarrow m_{A O} \cdot m_{B C} = - 1$
$C O ⊥ A B \Rightarrow m_{C O} \cdot m_{A B} = - 1$

Slopes of $A O$ and $B C$

Slope of $A O$ ( $m_{A O}$ ):
$m_{A O} = \frac{- 1 - 0}{5 - 0} = - \frac{1}{5}$
Slope of $B C$ ( $m_{B C}$ ):
$m_{B C} = \frac{k - 3}{h - ( - 2 )} = \frac{k - 3}{h + 2}$

Forming the First Equation

Using $m_{A O} \cdot m_{B C} = - 1$ :
$(- \frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = - 1$
$k - 3 = 5 (h + 2)$
$k - 3 = 5 h + 10$
Equation 1: $5 h - k + 13 = 0$

Slopes of $A B$ and $C O$

Slope of $C O$ ( $m_{C O}$ ):
$m_{C O} = \frac{k - 0}{h - 0} = \frac{k}{h}$
Slope of $A B$ ( $m_{A B}$ ):
$m_{A B} = \frac{3 - ( - 1 )}{- 2 - 5} = \frac{4}{- 7} = - \frac{4}{7}$

Forming the Second Equation

Using $m_{C O} \cdot m_{A B} = - 1$ :
$(\frac{k}{h}) \cdot (- \frac{4}{7}) = - 1$
$\frac{- 4 k}{7 h} = - 1$
$4 k = 7 h$
Equation 2: $7 h - 4 k = 0$

Solving for Vertex $C$

From Equation 2: $k = \frac{7 h}{4}$
Substitute in Equation 1:
$5 h - \frac{7 h}{4} + 13 = 0$
Multiply by 4: $20 h - 7 h + 52 = 0$
$13 h = - 52 \Rightarrow h = - 4$
Find $k$ : $k = \frac{7 ( - 4 )}{4} = - 7$
Vertex $C$ is $(- 4, - 7)$ . Statement $S 1$ is correct.

Introduction to Statement $S 2$

Statement $S 2$ :
Condition: $2 a, b, c$ are three consecutive terms of an A.P.
Line equation: $a x + b y + c = 0$
Claim: These lines are concurrent at $(2, - 2)$

The A.P. Condition

Since $2 a, b, c$ are in A.P.:
Common difference is constant: $b - 2 a = c - b$
$2 b = 2 a + c$
Rearranging to match the form $a x + b y + c = 0$ :
$2 a - 2 b + c = 0$

Identifying the Point of Concurrency

Compare the derived condition with the line equation:
1. $a x + b y + c = 0$
2. $a (2) + b (- 2) + c = 0$
By direct comparison, we can see that $x = 2$ and $y = - 2$ .
Therefore, the lines always pass through the fixed point $(2, - 2)$ .
Statement $S 2$ is correct.

Final Conclusion

Statement $S 1$ is correct.
Statement $S 2$ is correct.
Final Answer: Both statements are correct.

00:00 / 00:00

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Solution Diagram

Analyzing the Geometry of Orthocenters

The orthocenter

O (0, 0)

is the point of concurrency for the altitudes of a triangle. Given vertices

A (5, - 1)

and

B (- 2, 3)

, we define the third vertex as

C (h, k)

The altitude from

A

B C

implies that the line

A O

is perpendicular to

B C

. The slope of

A O

is calculated as:

m_{A O} = \frac{- 1 - 0}{5 - 0} = - \frac{1}{5}

The slope of

B C

is given by:

m_{B C} = \frac{k - 3}{h + 2}

Since

A O ⊥ B C

, the product of their slopes must be

- 1

. This yields the equation:

(- \frac{1}{5}) \cdot (\frac{k - 3}{h + 2}) = - 1 \Rightarrow k - 3 = 5 (h + 2) \Rightarrow 5 h - k + 13 = 0

Solving for the Third Vertex

Next, we consider the altitude from

C

A B

, implying

C O ⊥ A B

. The slope of

C O

m_{C O} = \frac{k}{h}

, and the slope of

A B

is:

m_{A B} = \frac{3 - ( - 1 )}{- 2 - 5} = - \frac{4}{7}

Applying the perpendicularity condition

m_{C O} \cdot m_{A B} = - 1

(\frac{k}{h}) \cdot (- \frac{4}{7}) = - 1 \Rightarrow 4 k = 7 h

Substituting

k = \frac{7 h}{4}

into the first equation

5 h - k + 13 = 0

5 h - \frac{7 h}{4} + 13 = 0 \Rightarrow 20 h - 7 h + 52 = 0 \Rightarrow 13 h = - 52

Solving these gives

h = - 4

and

k = - 7

. Thus, the vertex

C

(- 4, - 7)

, confirming that Statement $S 1$ is correct.

The Beauty of Concurrency

We now examine the family of lines

a x + b y + c = 0

where

2 a, b, c

are in Arithmetic Progression. The condition for an A.P. is

b - 2 a = c - b

, which simplifies to:

2 b = 2 a + c \Rightarrow 2 a - 2 b + c = 0

We compare this to the general line equation

a x + b y + c = 0

. By inspection, the equation holds true for all

a, b, c

if:

x = 2, y = - 2

This confirms that all such lines pass through the fixed point

(2, - 2)

. Therefore, Statement $S 2$ is also correct.

Common Traps & FAQs

Why does the orthocenter being at the origin make the math easier?

It simplifies the slope calculation significantly. The slope of the line from

(0, 0)

(x, y)

is simply

y / x

. This eliminates constant terms in your slope expressions, making the algebra much cleaner and less prone to errors.

How do I know the lines are concurrent at

(2, - 2)

The equation

a x + b y + c = 0

represents a family of lines. If you can express this as

a (2) + b (- 2) + c = 0

, you have essentially shown that the point

(2, - 2)

satisfies the equation for *any*

a, b, c

that satisfy the A.P. condition. It is a beautiful way to prove concurrency without finding the intersection of two specific lines.

Is it possible for the orthocenter to be outside the triangle?

Absolutely! If the triangle is obtuse, the orthocenter lies outside. Don't let the visual of the orthocenter being 'inside' restrict your algebraic approach. The slope property

m_{1} \cdot m_{2} = - 1

holds true regardless of the triangle's type.

Select Answer:

Visualized Solution

Introduction to Statement $S 1$

The Orthocenter Property

Slopes of $A O$ and $B C$

Forming the First Equation

Slopes of $A B$ and $C O$

Forming the Second Equation

Solving for Vertex $C$

Introduction to Statement $S 2$

The A.P. Condition

Identifying the Point of Concurrency

Final Conclusion

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Geometry of Orthocenters

Solving for the Third Vertex

The Beauty of Concurrency

Similar Questions

(a) Two vertices of a triangle are $(5, - 1)$ and $(- 2, 3)$ . If the orthocentre of the triangle is the origin, find the coordinates of the third point. (b) Find the equation of the line which bisects the obtuse angle between the lines $x - 2 y + 4 = 0$ and $4 x - 3 y + 2 = 0$ .

If a $Δ A B C$ has vertices $A (- 1, 7)$ , $B (- 7, 1)$ and $C (5, - 5)$ , then its orthocentre has coordinates

The vertices of a triangle are $[a t_{1} t_{2}, a (t_{1} + t_{2})]$ , $[a t_{2} t_{3}, a (t_{2} + t_{3})]$ , $[a t_{3} t_{1}, a (t_{3} + t_{1})]$ . Find the orthocentre of the triangle.

Two vertices of a triangle are $(0, 2)$ and $(4, 3)$ . If its orthocentre is at the origin, then its third vertex lies in which quadrant ?

Let the position vectors of three vertices of a triangle be $4 p + q - 3 r$ , $- 5 p + q + 2 r$ and $2 p - q + 2 r$ If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{p + q + r}{4}$ and $α p + β q + γ r$ respectively, then $α + 2 β + 5 γ$ is equal to:

The equations of the sides $A B, B C$ and $C A$ of a triangle $A B C$ are $2 x + y = 0$ , $x + p y = 15 a$ and $x - y = 3$ respectively. If its orthocentre is $(2, a)$ , $- \frac{1}{2} < a < 2$ , then $p$ is equal to

If $(α, β)$ is the orthocenter of the triangle $A B C$ with vertices $A (3, - 7), B (- 1, 2)$ and $C (4, 5)$ , then $9 α - 6 β + 60$ is equal to

Two vertices of a triangle $A B C$ are $A (3, - 1)$ and $B (- 2, 3)$ , and its orthocentre is $P (1, 1)$ . If the coordinates of the point $C$ are $(α, β)$ and the centre of the circle circumscribing the triangle $P A B$ is $(h, k)$ , then the value of $(α + β) + 2 (h + k)$ equals

The equations of the sides $A B, B C$ and $C A$ of a triangle $A B C$ are $2 x + y = 0, x + p y = 39$ and $x - y = 3$ respectively and $P (2, 3)$ is its circumcentre. Then which of the following is NOT true :

Let $A B C$ be the triangle such that the equations of lines $A B$ and $A C$ be $3 y - x = 2$ and $x + y = 2$ , respectively, and the points $B$ and $C$ lie on $x$ -axis. If $P$ is the orthocentre of the triangle $A B C$ , then the area of the triangle $P B C$ is equal to

Select Answer:

Visualized Solution

Introduction to Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S1

The Orthocenter Property

Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AO and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

Forming the First Equation

Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AB and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } CO

Forming the Second Equation

Solving for Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Introduction to Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S2

The A.P. Condition

Identifying the Point of Concurrency

Final Conclusion

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

Analyzing the Geometry of Orthocenters

Solving for the Third Vertex

The Beauty of Concurrency

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a ΔABC has vertices A(−1,7), B(−7,1) and C(5,−5), then its orthocentre has coordinates

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vertices of a triangle are [at1​t2​,a(t1​+t2​)], [at2​t3​,a(t2​+t3​)], [at3​t1​,a(t3​+t1​)]. Find the orthocentre of the triangle.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Two vertices of a triangle are (0,2) and (4,3). If its orthocentre is at the origin, then its third vertex lies in which quadrant ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equations of the sides AB,BC and CA of a triangle ABC are 2x+y=0, x+py=15a and x−y=3 respectively. If its orthocentre is (2,a), −21​<a<2, then p is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If (α,β) is the orthocenter of the triangle ABC with vertices A(3,−7),B(−1,2) and C(4,5), then 9α−6β+60 is equal to

Introduction to Statement $S 1$

Slopes of $A O$ and $B C$

Slopes of $A B$ and $C O$

Solving for Vertex $C$

Introduction to Statement $S 2$

If a $Δ A B C$ has vertices $A (- 1, 7)$ , $B (- 7, 1)$ and $C (5, - 5)$ , then its orthocentre has coordinates

The vertices of a triangle are $[a t_{1} t_{2}, a (t_{1} + t_{2})]$ , $[a t_{2} t_{3}, a (t_{2} + t_{3})]$ , $[a t_{3} t_{1}, a (t_{3} + t_{1})]$ . Find the orthocentre of the triangle.

Two vertices of a triangle are $(0, 2)$ and $(4, 3)$ . If its orthocentre is at the origin, then its third vertex lies in which quadrant ?

The equations of the sides $A B, B C$ and $C A$ of a triangle $A B C$ are $2 x + y = 0$ , $x + p y = 15 a$ and $x - y = 3$ respectively. If its orthocentre is $(2, a)$ , $- \frac{1}{2} < a < 2$ , then $p$ is equal to

If $(α, β)$ is the orthocenter of the triangle $A B C$ with vertices $A (3, - 7), B (- 1, 2)$ and $C (4, 5)$ , then $9 α - 6 β + 60$ is equal to