JEE Main 2019 (12 April)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3). Then the centroid of this triangle is :

Select Answer:

Progressive Hints

Hint 1

Visualize the geometry: You are given vertex

A (1, 2)

and the midpoints

M_{1} (- 1, 1)

and

M_{2} (2, 3)

of the sides

A B

and

A C

respectively. Think of the midpoint formula as a bridge between the known vertex and the unknown vertices

B

and

C

. How can you rearrange

M = \frac{A + B}{2}

to isolate

B

Hint 2

Apply the vector-like property of midpoints: If

M

is the midpoint of

A B

, then

B = 2 M - A

. This simple algebraic manipulation allows you to find the coordinates of

B

and

C

directly without needing to solve complex systems of equations.

Hint 3

Once you have determined the coordinates of all three vertices

A (1, 2)

B (x_{B}, y_{B})

, and

C (x_{C}, y_{C})

, the centroid

G

is simply the arithmetic mean of the coordinates:

G = (\frac{x _{A} + x _{B} + x _{C}}{3}, \frac{y _{A} + y _{B} + y _{C}}{3})

. Just plug in your calculated values and simplify.

Visualized Solution

Visualizing the Given Information

Given Vertex: $A (1, 2)$
Midpoints: $M_{1} (- 1, 1)$ and $M_{2} (2, 3)$

The Strategy

Goal: Find vertices $B$ and $C$ to calculate the centroid.

The Midpoint Formula

Midpoint Formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})$
Let $B = (x_{B}, y_{B})$ and $C = (x_{C}, y_{C})$

Setting up for Vertex $B$

For side $A B$ , $M_{1} (- 1, 1)$ is the midpoint.
$\frac{1 + x _{B}}{2} = - 1$
$\frac{2 + y _{B}}{2} = 1$

Calculating Vertex $B$

$1 + x_{B} = - 2 ⟹ x_{B} = - 3$
$2 + y_{B} = 2 ⟹ y_{B} = 0$
Vertex $B = (- 3, 0)$

Setting up for Vertex $C$

For side $A C$ , $M_{2} (2, 3)$ is the midpoint.
$\frac{1 + x _{C}}{2} = 2$
$\frac{2 + y _{C}}{2} = 3$

Calculating Vertex $C$

$1 + x_{C} = 4 ⟹ x_{C} = 3$
$2 + y_{C} = 6 ⟹ y_{C} = 4$
Vertex $C = (3, 4)$

Completing the Triangle

Vertices found:
$A (1, 2)$
$B (- 3, 0)$
$C (3, 4)$

The Centroid Formula

Centroid Formula: $G = (\frac{x _{A} + x _{B} + x _{C}}{3}, \frac{y _{A} + y _{B} + y _{C}}{3})$

Substituting the Vertices

$G_{x} = \frac{1 + ( - 3 ) + 3}{3}$
$G_{y} = \frac{2 + 0 + 4}{3}$

Final Calculation

$G_{x} = \frac{1}{3}$
$G_{y} = \frac{6}{3} = 2$
Centroid $G = (\frac{1}{3}, 2)$

00:00 / 00:00

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

Solution Diagram

Analyzing the Setup

To find the centroid of the triangle, we must first determine the coordinates of the vertices

B

and

C

. We are given vertex

A (1, 2)

and the midpoints of the sides

A B

and

A C

, which are

M_{1} (- 1, 1)

and

M_{2} (2, 3)

respectively.

The midpoint formula states that the midpoint of a segment with endpoints

(x_{1}, y_{1})

and

(x_{2}, y_{2})

is given by:

M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

Finding the Missing Vertices

Since

M_{1}

is the midpoint of

A B

, we can express

B

B = 2 M_{1} - A

. Substituting the given coordinates:

x_{B} = 2 (- 1) - 1 = - 3

y_{B} = 2 (1) - 2 = 0

Thus, vertex

B

(- 3, 0)

. Similarly, for vertex

C

using midpoint

M_{2}

x_{C} = 2 (2) - 1 = 3

y_{C} = 2 (3) - 2 = 4

Thus, vertex

C

(3, 4)

Calculating the Centroid

The centroid

G

of a triangle with vertices

(x_{1}, y_{1})

(x_{2}, y_{2})

, and

(x_{3}, y_{3})

is the arithmetic mean of the coordinates:

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

Substituting the vertices

A (1, 2)

B (- 3, 0)

, and

C (3, 4)

into the formula:

G = (\frac{1 - 3 + 3}{3}, \frac{2 + 0 + 4}{3})

G = (\frac{1}{3}, \frac{6}{3})

The final coordinates of the centroid are $(\frac{1}{3}, 2)$ .

Common Traps & FAQs

Wait, why do we reject the negative root here? Or is there no quadratic involved?

Great observation! In this specific problem, there is no quadratic equation involved because the midpoint formula is linear. We are simply dealing with basic arithmetic. If you find yourself forming a quadratic, you might be overcomplicating the geometry. Stick to the linear relationship

B = 2 M - A

Why can't we just use energy conservation or some other physics principle?

While I love the enthusiasm for physics, this is a pure coordinate geometry problem! We are working in the realm of Euclidean space where the properties of triangles are defined by the positions of their vertices. Keep your physics toolkit for mechanics and electromagnetism, and use your coordinate geometry toolkit here.

Is there a shortcut using the midpoints directly to find the centroid?

Yes, there is! The centroid of the triangle formed by the midpoints is actually the same as the centroid of the original triangle. You could calculate the centroid of the triangle with vertices

A (1, 2)

M_{1} (- 1, 1)

, and

M_{2} (2, 3)

and get the same result. It is a beautiful property of triangles, but finding the vertices

B

and

C

first is a more fundamental way to understand the structure.

Animated Solution for Mathematics - Straight Lines: A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3). Then the centroid of this triangle is :

Select Answer:

Visualized Solution

Visualizing the Given Information

The Strategy

The Midpoint Formula

Setting up for Vertex $B$

Calculating Vertex $B$

Setting up for Vertex $C$

Calculating Vertex $C$

Completing the Triangle

The Centroid Formula

Substituting the Vertices

Final Calculation

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

Analyzing the Setup

Finding the Missing Vertices

Calculating the Centroid

Similar Questions

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Animated Solution for Mathematics - Straight Lines: A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3). Then the centroid of this triangle is :

Select Answer:

Visualized Solution

Visualizing the Given Information

The Strategy

The Midpoint Formula

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

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

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

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

Completing the Triangle

The Centroid Formula

Substituting the Vertices

Final Calculation

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

Analyzing the Setup

Finding the Missing Vertices

Calculating the Centroid

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the vertices P,Q,R of a triangle PQR are rational points, which of the following points of the triangle PQR is (are) always rational point(s)?

Setting up for Vertex $B$

Calculating Vertex $B$

Setting up for Vertex $C$

Calculating Vertex $C$

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If the vertices $P, Q, R$ of a triangle $P QR$ are rational points, which of the following points of the triangle $P QR$ is (are) always rational point(s)?