How to find the third vertex given orthocentre?

Use perpendicularity conditions: AH⊥BC and BH⊥AC. Set up two equations using dot product = 0 and solve simultaneously.

What is the orthocentre of a triangle?

The orthocentre is the point where all three altitudes of a triangle intersect. An altitude is perpendicular to the opposite side.

With H=(0,0), AH⊥BC gives 5h-k=-13. BH⊥AC gives -2h+3k=-13. Solving: h=-4, k=-7. Third vertex is (-4,-7) ✓

What does 2a, b, c in AP mean?

Three terms are in AP means common difference is equal: b-2a = c-b, so 2b = 2a+c, giving c = 2b-2a.

How to find point of concurrence?

Substitute AP condition c=2b-2a into ax+by+c=0: a(x-2)+b(y+2)=0. This family always passes through x=2, y=-2.

Why does a(x-2)+b(y+2)=0 give point (2,-2)?

For any values of a,b, x=2 and y=-2 satisfy a(2-2)+b(-2+2)=0+0=0. So all lines pass through (2,-2).

How to check if two lines are perpendicular?

Two lines are perpendicular if their direction vectors have zero dot product, or equivalently m1·m2 = -1 for their slopes.

What is the condition for concurrent lines?

Lines are concurrent if they all pass through a single common point. To show this, demonstrate one point satisfies all equations.

How to solve the simultaneous equations in S1?

From 5h-k=-13: k=5h+13. Substitute into -2h+3k=-13: 13h=-52, h=-4. Then k=-7.

Is the point (2,-2) special for AP lines?

Yes! For any a,b with 2a,b,c in AP, the line ax+by+c=0 always passes through (2,-2). This is the fixed point of the family.

Among statements S1 orthocentre third vertex S2 AP concurrent lines

Among statements S1 orthocentre third vertex S2 AP concurrent lines | JEE Main Mathematics

NEETJEE

Math Geometry AP

Q MCQ Statements

Among the statements:

STATEMENT 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 $(-4,-7)$.

STATEMENT S2

If positive numbers $2a, b, c$ are three consecutive terms of an A.P., then the lines $ax + by + c = 0$ are concurrent at $(2,-2)$.

A) both are incorrect

B) only (S2) is correct

C) both are correct ✓

D) only (S1) is correct

✅ Correct Answer

C) Both are correct

✓

Solution Steps

▶ Verifying S1

Set up perpendicularity conditions

Let third vertex be $C(h,k)$. Orthocentre $H(0,0)$.

Condition 1: $AH \perp BC$

$\vec{AH} = (0-5,\;0+1) = (-5,\;1)$

$\vec{BC} = (h+2,\;k-3)$

Dot product $= 0$: $(-5)(h+2) + (1)(k-3) = 0$

$$-5h – 10 + k – 3 = 0 \implies \boxed{5h – k = -13} \quad \cdots(i)$$

Second perpendicularity condition

Condition 2: $BH \perp AC$

$\vec{BH} = (0+2,\;0-3) = (2,\;-3)$

$\vec{AC} = (h-5,\;k+1)$

Dot product $= 0$: $2(h-5) + (-3)(k+1) = 0$

$$2h – 10 – 3k – 3 = 0 \implies \boxed{2h – 3k = 13} \quad \cdots(ii)$$

Solve simultaneous equations

From (i): $k = 5h + 13$

Substitute into (ii):

$2h – 3(5h + 13) = 13$

$2h – 15h – 39 = 13$

$-13h = 52 \implies h = -4$

$k = 5(-4) + 13 = -7$

∴ Third vertex = $(-4,-7)$ ✓ — S1 is CORRECT

▶ Verifying S2

Use AP condition

$2a, b, c$ are in A.P. $\Rightarrow$ common difference is equal:

$b – 2a = c – b \implies 2b = 2a + c$

$$\therefore c = 2b – 2a \quad \cdots(*)$$

Substitute into line equation

Line: $ax + by + c = 0$

Replace $c$ using $(*)$:

$ax + by + (2b – 2a) = 0$

$ax – 2a + by + 2b = 0$

$$a(x – 2) + b(y + 2) = 0$$

Identify fixed point of concurrence

$a(x-2) + b(y+2) = 0$ holds for ALL values of $a,b$ if and only if:

$$x – 2 = 0 \quad \text{AND} \quad y + 2 = 0$$

$$\implies x = 2,\quad y = -2$$

∴ All lines pass through $(2,-2)$ ✓ — S2 is CORRECT

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Key Insight

Both S1 and S2 are correct → Answer C. AP condition always reduces to a(x−2)+b(y+2)=0, a fixed point family.

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Theory

1. Orthocentre and Altitude Perpendicularity

The orthocentre $H$ of a triangle is the intersection of its three altitudes. An altitude from vertex $A$ is perpendicular to side $BC$. If $H$ is known, then for any vertex $A$: $\vec{AH} \perp \vec{BC}$, giving $\vec{AH} \cdot \vec{BC} = 0$. Two such conditions from two known vertices yield two equations to find the unknown third vertex.

2. Arithmetic Progression Condition

Three numbers $p, q, r$ are in A.P. iff $q – p = r – q$, i.e., $2q = p + r$. For $2a, b, c$: $2b = 2a + c$, giving $c = 2b – 2a$. This linear relation between $a, b, c$ is the key constraint that determines which point all family lines $ax+by+c=0$ pass through.

3. Family of Lines Through a Fixed Point

The equation $\lambda\,L_1 + \mu\,L_2 = 0$ represents a family of lines passing through the intersection of $L_1=0$ and $L_2=0$, for all values of $\lambda, \mu$. Here $a(x-2) + b(y+2) = 0$ is exactly such a family (with $L_1: x-2=0$, $L_2: y+2=0$), concurrent at $(2,-2)$.

4. Dot Product for Perpendicularity

Two vectors $\mathbf{u} = (u_1, u_2)$ and $\mathbf{v} = (v_1, v_2)$ are perpendicular iff $\mathbf{u}\cdot\mathbf{v} = u_1v_1 + u_2v_2 = 0$. This is equivalent to their slopes satisfying $m_1 m_2 = -1$. In orthocentre problems, always form direction vectors from the known point to unknown vertex, then apply the dot product condition.

5. Verifying Statement-Type Questions

For “Among the statements” JEE problems: verify each statement independently with rigorous calculation. Don’t guess based on patterns. S1 needed two dot product equations and solving 2×2 linear system. S2 needed the AP substitution trick to reveal the fixed point structure. Both verified true independently → answer C.

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FAQs

How to find third vertex given orthocentre?

⌄

Use AH⊥BC and BH⊥AC. Form dot product = 0 for each. Solve the 2 equations for (h,k).

What is the equation from AH⊥BC?

⌄

AH = (-5,1), BC = (h+2,k-3). Dot product: -5(h+2)+(k-3)=0 → 5h-k = -13.

What does 2a,b,c in AP give us?

⌄

2b = 2a+c so c = 2b-2a. Substituting: ax+by+(2b-2a) = a(x-2)+b(y+2) = 0.

Why is a(x-2)+b(y+2)=0 concurrent at (2,-2)?

⌄

For ANY a,b: substituting x=2,y=-2 gives a(0)+b(0)=0. So (2,-2) lies on every line of this family.

Is the answer C (both correct)?

⌄

Yes. S1: third vertex = (-4,-7) verified by perpendicularity. S2: concurrent at (2,-2) verified by AP substitution. Both TRUE → C.

What is the slope method alternative for S1?

⌄

slope(AH) = (0-(-1))/(0-5) = -1/5. Altitude AH ⊥ BC → slope(BC) = 5. slope(BH) = (0-3)/(0+2) = -3/2 → slope(AC) = 2/3. Find C as intersection of lines through A,B with these slopes.

How to solve 5h−k=−13 and 2h−3k=13?

⌄

From eq(i): k=5h+13. Put in eq(ii): 2h-3(5h+13)=13 → -13h=52 → h=-4. Then k=-7.

What is a family of lines?

⌄

A set of lines depending on parameters. λL₁+μL₂=0 all pass through intersection of L₁=L₂=0. Here a(x-2)+b(y+2)=0 passes through (2,-2) for all a,b.

Do we need to verify with third altitude in S1?

⌄

No. Two conditions from two vertices uniquely determine C(h,k). The third altitude condition is automatically satisfied (it’s redundant for a valid triangle).

What if 2a,b,c were in GP instead of AP?

⌄

In GP: b²=2ac. That’s a non-linear constraint and doesn’t directly give a linear fixed-point. The AP condition is special in giving the linear relation c=2b-2a.

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