JEE Advanced 2009

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Animated Solution for Mathematics - Straight Lines: The locus of the orthocentre of the triangle formed by the lines $(1 + p) x - p y + p (1 + p) = 0$ , $(1 + q) x - q y + q (1 + q) = 0$ , and $y = 0$ , where $p \neq = q$ , is

Q: I got $h = pq$ and $k = -pq$. How do I know the locus is a straight line?

Look at the relationship between $h$ and $k$. We have $h = pq$ and $k = -pq$. If you add them, you get $h + k = 0$. This is a linear equation in terms of the coordinates of the orthocenter. Since the locus is the set of all points $(h, k)$ that satisfy this condition, replacing $(h, k)$ with $(x, y)$ gives us $x + y = 0$, which is the equation of a straight line. It is beautiful, isn't it?

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the geometry. The triangle has one side lying on the x-axis (

y = 0

). This is a massive advantage! Think about what this implies for the altitude dropped from the vertex opposite to the x-axis. It must be a vertical line, right? That simplifies one of your two required equations immediately.

Hint 2

Hint 2: To find the orthocenter, you need the intersection of two altitudes. You have the first one (the vertical line). For the second altitude, you need the slope of the side

A C

. Remember, the altitude from vertex

B

is perpendicular to

A C

. Use the negative reciprocal of the slope of

A C

to find the slope of this altitude.

Hint 3

Hint 3: You need the coordinates of vertex

A

. Solve the system of equations for

L_{1}

and

L_{2}

to find

A (pq, (p + 1) (q + 1))

. Once you have

A

and

B (- p, 0)

, calculate the slope of

A C

. Then, construct the equation of the altitude from

B

using the point-slope form and intersect it with the vertical line

x = pq

Visualized Solution

The Triangle Setup

Given lines forming the triangle:
$L_{1} : (1 + p) x - p y + p (1 + p) = 0$
$L_{2} : (1 + q) x - q y + q (1 + q) = 0$
$L_{3} : y = 0$ (the x-axis)
We need to find the locus of the orthocenter $H (h, k)$ .

Finding Vertices on the x-axis

Since $L_{3}$ is $y = 0$ , we find its intersection with $L_{1}$ and $L_{2}$ .
For $L_{1}$ , put $y = 0$ : $(1 + p) x + p (1 + p) = 0$
$x = - p ⟹ B (- p, 0)$
For $L_{2}$ , put $y = 0$ : $(1 + q) x + q (1 + q) = 0$
$x = - q ⟹ C (- q, 0)$

Finding Vertex $A$

Vertex $A$ is the intersection of $L_{1}$ and $L_{2}$ .
Rewrite $L_{1}$ : $y = \frac{1 + p}{p} x + (1 + p)$
Rewrite $L_{2}$ : $y = \frac{1 + q}{q} x + (1 + q)$
Equating $y$ : $\frac{1 + p}{p} x + 1 + p = \frac{1 + q}{q} x + 1 + q$

Solving for $x$ -coordinate of $A$

Rearranging terms: $x (\frac{1 + p}{p} - \frac{1 + q}{q}) = q - p$
$x (\frac{1}{p} + 1 - \frac{1}{q} - 1) = q - p$
$x (\frac{q - p}{pq}) = q - p$
Since $p \neq = q$ , we get $x = pq$ .

Solving for $y$ -coordinate of $A$

Substitute $x = pq$ into $L_{1}$ :
$y = \frac{1 + p}{p} (pq) + (1 + p)$
$y = q (1 + p) + (1 + p)$
$y = (p + 1) (q + 1)$
So, $A = (pq, (p + 1) (q + 1))$ .

The First Altitude

The orthocenter $H$ is the intersection of altitudes.
The altitude from $A$ is perpendicular to $B C$ .
Since $B C$ lies on the x-axis ( $y = 0$ ), the altitude from $A$ must be a vertical line.
Equation of vertical line through $A$ : $x = pq$ .

Slope of $A C$

To find the altitude from $B$ , we first need the slope of $A C$ .
$A = (pq, (p + 1) (q + 1))$ and $C = (- q, 0)$ .
$m_{A C} = \frac{( p + 1 ) ( q + 1 ) - 0}{pq - ( - q )}$
$m_{A C} = \frac{( p + 1 ) ( q + 1 )}{q ( p + 1 )} = \frac{q + 1}{q}$

The Second Altitude

The altitude from $B$ is perpendicular to $A C$ .
Its slope is $m_{⊥} = - \frac{1}{m _{A C}} = - \frac{q}{q + 1}$ .
It passes through $B (- p, 0)$ .
Equation: $y - 0 = - \frac{q}{q + 1} (x - (- p))$
$y = - \frac{q}{q + 1} (x + p)$

Finding the Orthocenter $H$

Let $H (h, k)$ be the orthocenter.
It lies on the first altitude: $h = pq$ .
Substitute $x = h = pq$ into the second altitude to find $k$ :
$k = - \frac{q}{q + 1} (pq + p)$

Simplifying the $y$ -coordinate of $H$

$k = - \frac{q}{q + 1} \cdot p (q + 1)$
Canceling $(q + 1)$ : $k = - pq$ .
So, the orthocenter is $H (pq, - pq)$ .

The Locus of $H$

We have parametric equations for $H$ :
$h = pq$
$k = - pq$
Adding them: $h + k = pq - pq = 0$ .
Replacing $(h, k)$ with $(x, y)$ , the locus is $x + y = 0$ .
This represents a straight line.

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We are given three lines that define a triangle:

L_{1} : (1 + p) x - p y + p (1 + p) = 0

L_{2} : (1 + q) x - q y + q (1 + q) = 0

L_{3} : y = 0

The line

L_{3}

represents the x-axis. To find the vertices

B

and

C

located on the x-axis, we set

y = 0

in the equations for

L_{1}

and

L_{2}

For

L_{1}

, we obtain

(1 + p) x + p (1 + p) = 0

, which simplifies to

x = - p

. Thus, the vertex is

B (- p, 0)

For

L_{2}

, we obtain

(1 + q) x + q (1 + q) = 0

, which simplifies to

x = - q

. Thus, the vertex is

C (- q, 0)

Finding the Third Vertex

To find vertex

A

, we determine the intersection of

L_{1}

and

L_{2}

. We express

y

in terms of

x

for both lines:

y = \frac{1 + p}{p} x + (1 + p)

y = \frac{1 + q}{q} x + (1 + q)

Equating these expressions, we have:

\frac{1 + p}{p} x + (1 + p) = \frac{1 + q}{q} x + (1 + q)

Rearranging the terms leads to:

x (\frac{1}{p} - \frac{1}{q}) = q - p

Applying the common denominator, we get

x (\frac{q - p}{pq}) = q - p

. Assuming $p eq q$, we cancel

(q - p)

to find

x = pq

. Substituting this into

L_{1}

yields the y-coordinate

y = (p + 1) (q + 1)

. Therefore,

A = (pq, (p + 1) (q + 1))

Determining the Orthocenter

The orthocenter

H (h, k)

is the intersection of the altitudes. The altitude from

A

to the base

B C

(the x-axis) is a vertical line defined by the x-coordinate of

A

Thus, our first relation is:

h = pq

To find the second altitude, we calculate the slope of

A C

using

A (pq, (p + 1) (q + 1))

and

C (- q, 0)

m_{A C} = \frac{( p + 1 ) ( q + 1 ) - 0}{pq - ( - q )} = \frac{( p + 1 ) ( q + 1 )}{q ( p + 1 )} = \frac{q + 1}{q}

The altitude from

B

is perpendicular to

A C

, so its slope is

m_{⊥} = - \frac{q}{q + 1}

The Final Reveal

Using the point-slope form for the altitude passing through

B (- p, 0)

, we write:

y - 0 = - \frac{q}{q + 1} (x + p)

Substituting the orthocenter's coordinates

(h, k)

where

h = pq

, we get:

k = - \frac{q}{q + 1} (pq + p)

Factoring

p

out of the term

(pq + p)

, we obtain:

k = - \frac{q}{q + 1} \cdot p (q + 1)

The

(q + 1)

terms cancel, leaving

k = - pq

. Since

h = pq

and

k = - pq

, we conclude that

h + k = 0

. Replacing

h

and

k

with

x

and

y

, the locus of the orthocenter is:

$x + y = 0$

Common Traps & FAQs

Wait, why is the altitude from vertex A just a vertical line? Can't it be slanted?

That is a fantastic question! Remember, an altitude is defined as a line segment from a vertex perpendicular to the opposite side. Here, the opposite side is the x-axis (

y = 0

). Any line perpendicular to a horizontal line must be a vertical line. Since the altitude passes through vertex

A

, its equation is simply

x = x_{A}

. It is not just a guess; it is a geometric necessity!

Why do we need to find the intersection of

L_{1}

and

L_{2}

to get vertex A? Can't we just use the orthocenter formula?

You could, but the general formula for the orthocenter is often a trap in competitive exams—it is algebraically heavy and prone to calculation errors. By finding vertex

A

explicitly, we keep the algebra clean and manageable. In JEE Advanced, the 'brute force' method is often less elegant than the 'geometric insight' method. Trust the process of finding the vertices first!

I got

h = pq

and

k = - pq

. How do I know the locus is a straight line?

Look at the relationship between

h

and

k

. We have

h = pq

and

k = - pq

. If you add them, you get

h + k = 0

. This is a linear equation in terms of the coordinates of the orthocenter. Since the locus is the set of all points

(h, k)

that satisfy this condition, replacing

(h, k)

with

(x, y)

gives us

x + y = 0

, which is the equation of a straight line. It is beautiful, isn't it?

Animated Solution for Mathematics - Straight Lines: The locus of the orthocentre of the triangle formed by the lines $(1 + p) x - p y + p (1 + p) = 0$ , $(1 + q) x - q y + q (1 + q) = 0$ , and $y = 0$ , where $p \neq = q$ , is

Select Answer:

Visualized Solution

The Triangle Setup

Finding Vertices on the x-axis

Finding Vertex $A$

Solving for $x$ -coordinate of $A$

Solving for $y$ -coordinate of $A$

The First Altitude

Slope of $A C$

The Second Altitude

Finding the Orthocenter $H$

Simplifying the $y$ -coordinate of $H$

The Locus of $H$

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

Analyzing the Setup

Finding the Third Vertex

Determining the Orthocenter

The Final Reveal

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Select Answer:

Visualized Solution

The Triangle Setup

Finding Vertices on the x-axis

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

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

The First Altitude

Slope of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

The Second Altitude

Finding the Orthocenter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y-coordinate of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

The Locus of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } H

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

Analyzing the Setup

Finding the Third Vertex

Determining the Orthocenter

The Final Reveal

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