JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ and whose orthocenter is $(\frac{7}{3}, \frac{7}{3})$ is:

Select Answer:

Progressive Hints

Hint 1

Visualize the triangle as a geometric puzzle. You are given two sides, which means you know the slopes of two lines. The orthocenter is the intersection of altitudes. Remember, an altitude is simply a line perpendicular to a side, passing through the opposite vertex. This is your gateway to finding the missing vertices.

Hint 2

Your toolkit must include the point-slope form of a line:

y - y_{1} = m (x - x_{1})

. Since you know the slope of a side, you can easily find the slope of the altitude perpendicular to it. The orthocenter

H

is the common point for these altitudes, acting as the anchor for your calculations.

Hint 3

Start by finding vertex

A

by solving the intersection of the two given lines. Then, construct the equation of the altitude from

B

(perpendicular to side

A C

) and the altitude from

C

(perpendicular to side

A B

). Intersect these altitudes with the respective sides to find vertices

B

and

C

. Once you have all three vertices, the centroid is just the average of the coordinates.

Visualized Solution

Visualizing the Triangle Geometry

Given sides: $L_{1} : x - 2 y + 1 = 0$ and $L_{2} : 2 x - y - 1 = 0$
Orthocenter $H : (\frac{7}{3}, \frac{7}{3})$
Objective: Find centroid $G$ and distance $O G$ .

Finding Vertex $A$

Vertex $A$ is the intersection of $L_{1}$ and $L_{2}$ .
Solving: $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ .
$2 (2 y - 1) - y - 1 = 0 \Rightarrow 3 y - 3 = 0 \Rightarrow y = 1$ .
Substituting $y = 1$ : $x - 2 (1) + 1 = 0 \Rightarrow x = 1$ .
Vertex $A = (1, 1)$ .

The Orthocenter Property

Property: Altitude from a vertex is perpendicular to the opposite side.
All altitudes pass through the orthocenter $H (\frac{7}{3}, \frac{7}{3})$ .
Altitude from $B ⊥ L_{2}$ (side $A C$ ).
Altitude from $C ⊥ L_{1}$ (side $A B$ ).

Equation of Altitude from $B$

Let $L_{2}$ be side $A C : 2 x - y - 1 = 0$ . Its slope is $m_{A C} = 2$ .
Altitude from $B$ is perpendicular to $A C$ , so $m_{B H} = - \frac{1}{2}$ .
Equation of $B H$ : $y - \frac{7}{3} = - \frac{1}{2} (x - \frac{7}{3})$ .
Simplifying: $2 y - \frac{14}{3} = - x + \frac{7}{3} \Rightarrow x + 2 y = 7$ .

Finding Vertex $B$

Vertex $B$ lies on side $A B$ ( $L_{1} : x - 2 y + 1 = 0$ ) and altitude $B H$ ( $x + 2 y = 7$ ).
Adding equations: $(x - 2 y) + (x + 2 y) = - 1 + 7 \Rightarrow 2 x = 6 \Rightarrow x = 3$ .
Substituting $x = 3$ in $B H$ : $3 + 2 y = 7 \Rightarrow 2 y = 4 \Rightarrow y = 2$ .
Vertex $B = (3, 2)$ .

Equation of Altitude from $C$

Let $L_{1}$ be side $A B : x - 2 y + 1 = 0$ . Its slope is $m_{A B} = \frac{1}{2}$ .
Altitude from $C$ is perpendicular to $A B$ , so $m_{C H} = - 2$ .
Equation of $C H$ : $y - \frac{7}{3} = - 2 (x - \frac{7}{3})$ .
Simplifying: $y - \frac{7}{3} = - 2 x + \frac{14}{3} \Rightarrow 2 x + y = 7$ .

Finding Vertex $C$

Vertex $C$ lies on side $A C$ ( $L_{2} : 2 x - y - 1 = 0$ ) and altitude $C H$ ( $2 x + y = 7$ ).
Adding equations: $(2 x - y) + (2 x + y) = 1 + 7 \Rightarrow 4 x = 8 \Rightarrow x = 2$ .
Substituting $x = 2$ in $C H$ : $2 (2) + y = 7 \Rightarrow y = 3$ .
Vertex $C = (2, 3)$ .

Calculating the Centroid $G$

Centroid formula: $G (x, y) = (\frac{x _{A} + x _{B} + x _{C}}{3}, \frac{y _{A} + y _{B} + y _{C}}{3})$
Vertices: $A (1, 1)$ , $B (3, 2)$ , $C (2, 3)$ .
Calculation: $G = (\frac{1 + 3 + 2}{3}, \frac{1 + 2 + 3}{3}) = (\frac{6}{3}, \frac{6}{3})$ .
Centroid $G = (2, 2)$ .

Final Distance from Origin

Origin $O = (0, 0)$ , Centroid $G = (2, 2)$ .
Distance formula: $d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$
$O G = (2 - 0)^{2} + (2 - 0)^{2}$
$O G = 4 + 4 = 8 = 22$ .

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

We begin by identifying the first vertex,

A

, which is the intersection of the two given lines:

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

L_{2} : 2 x - y - 1 = 0

To find

A

, we express

x

from

L_{1}

x = 2 y - 1

and substitute it into

L_{2}

2 (2 y - 1) - y - 1 = 0

Simplifying this yields

3 y - 3 = 0

, which gives

y = 1

. Substituting back, we find

x = 1

. Thus, our first anchor point is $A (1, 1)$ .

The Orthocenter's Secret

The orthocenter

H

is located at

(\frac{7}{3}, \frac{7}{3})

. Recall that an altitude from a vertex is perpendicular to the opposite side.

To find vertex

B

, we note it lies on

L_{1}

. The altitude from

B

must be perpendicular to

L_{2}

. Since the slope of

L_{2}

2

, the altitude slope is

- \frac{1}{2}

Using the point-slope form with

H

, we derive the altitude equation:

y - \frac{7}{3} = - \frac{1}{2} (x - \frac{7}{3})

Multiplying by

6

and rearranging, we obtain the equation:

x + 2 y = 7

The Hunt for Vertices

Vertex

B

is the intersection of

L_{1}

(

x - 2 y + 1 = 0

) and the altitude

x + 2 y = 7

. Adding these equations:

(x - 2 y) + (x + 2 y) = - 1 + 7

2 x = 6 \Rightarrow x = 3

Substituting

x = 3

into

x + 2 y = 7

, we find

y = 2

. Thus, $B = (3, 2)$ .

For vertex

C

, the altitude is perpendicular to

L_{1}

(slope

\frac{1}{2}

), so its slope is

- 2

. Using

H

, the equation is:

y - \frac{7}{3} = - 2 (x - \frac{7}{3}) \Rightarrow 2 x + y = 7

Vertex

C

is the intersection of

L_{2}

(

2 x - y - 1 = 0

) and

2 x + y = 7

. Adding these gives

4 x = 8

, so

x = 2

. Substituting back, we find

y = 3

. Thus, $C = (2, 3)$ .

The Centroid and the Final Distance

The centroid

G

is the arithmetic mean of the vertices

A (1, 1)

B (3, 2)

, and

C (2, 3)

G = (\frac{1 + 3 + 2}{3}, \frac{1 + 2 + 3}{3}) = (2, 2)

Finally, we calculate the distance of

G (2, 2)

from the origin

O (0, 0)

using the distance formula:

O G = (2 - 0)^{2} + (2 - 0)^{2} = 4 + 4 = 8

The final distance is $22$ .

Common Traps & FAQs

Why can't we just use the centroid formula immediately?

The centroid formula requires the coordinates of all three vertices:

A

B

, and

C

. We only have the intersection of two sides (Vertex

A

). We must first 'hunt' for vertices

B

and

C

using the orthocenter property. Think of it as a treasure hunt where you need to find the keys (vertices) before you can open the chest (the centroid).

How do I know which altitude corresponds to which side?

The altitude from a vertex is always perpendicular to the *opposite* side. If you are looking for the altitude from vertex

B

, it must be perpendicular to side

A C

(the line

L_{2}

). If you are looking for the altitude from vertex

C

, it must be perpendicular to side

A B

(the line

L_{1}

). Always keep track of which vertex is 'opposite' to which side.

Is there a shortcut using the Euler line?

While the Euler line connects the orthocenter, centroid, and circumcenter, it is not the most efficient path here. In coordinate geometry, when you are given the equations of the sides and the orthocenter, the most robust and 'JEE-proof' method is to construct the altitude equations and find the vertices directly. It minimizes the chance of conceptual errors.

Animated Solution for Mathematics - Straight Lines: The distance of the origin from the centroid of the triangle whose two sides have the equations $x - 2 y + 1 = 0$ and $2 x - y - 1 = 0$ and whose orthocenter is $(\frac{7}{3}, \frac{7}{3})$ is:

Select Answer:

Visualized Solution

Visualizing the Triangle Geometry

Finding Vertex $A$

The Orthocenter Property

Equation of Altitude from $B$

Finding Vertex $B$

Equation of Altitude from $C$

Finding Vertex $C$

Calculating the Centroid $G$

Final Distance from Origin

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

Analyzing the Setup

The Orthocenter's Secret

The Hunt for Vertices

The Centroid and the Final Distance

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Select Answer:

Visualized Solution

Visualizing the Triangle Geometry

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

The Orthocenter Property

Equation of Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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

Equation of Altitude from .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

Calculating the Centroid .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Final Distance from Origin

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

Analyzing the Setup

The Orthocenter's Secret

The Hunt for Vertices

The Centroid and the Final Distance

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