JEE Main 2020 - 9 Jan (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: $A (3, - 1)$ , $B (1, 3)$ , $C (2, 4)$ are vertices of $Δ A B C$ if $D$ is centroid of $Δ A B C$ and $P$ is point of intersection of lines $x + 3 y - 1 = 0$ and $3 x - y + 1 = 0$ then which of the following points lies on line joining $D$ and $P$

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

Vertices of the triangle are given as $A (3, - 1)$ , $B (1, 3)$ , and $C (2, 4)$ .
Our first goal is to locate the centroid, denoted by $D$ .

The Centroid Formula

The centroid $D$ of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , and $(x_{3}, y_{3})$ is the average of the coordinates.
Formula: $D = (\frac{x _{1} + x _{2} + x _{3}}{3}, \frac{y _{1} + y _{2} + y _{3}}{3})$

Substituting the Coordinates

Let's plug in the $x$ and $y$ coordinates of $A$ , $B$ , and $C$ .
$D = (\frac{3 + 1 + 2}{3}, \frac{- 1 + 3 + 4}{3})$

Calculating Centroid $D$

$x$ -coordinate: $\frac{6}{3} = 2$
$y$ -coordinate: $\frac{6}{3} = 2$
Therefore, the centroid is $D (2, 2)$ .

Introducing the Intersecting Lines

We are given two lines:
Line 1: $x + 3 y - 1 = 0 \Rightarrow x + 3 y = 1$
Line 2: $3 x - y + 1 = 0 \Rightarrow 3 x - y = - 1$
Their intersection point is $P$ .

Setting up Elimination

To find $P$ , we solve the system of linear equations.
Let's eliminate $y$ by multiplying Line 2 by $3$ .
Modified Line 2: $9 x - 3 y = - 3$

Solving for $x$

Add Line 1 and Modified Line 2:
$(x + 3 y) + (9 x - 3 y) = 1 + (- 3)$
$10 x = - 2$
$x = - \frac{1}{5}$

Solving for $y$

Substitute $x = - \frac{1}{5}$ back into Line 2: $3 x - y = - 1$
$3 (- \frac{1}{5}) - y = - 1$
$- y = - 1 + \frac{3}{5} = - \frac{2}{5}$
$y = \frac{2}{5}$
So, $P = (- \frac{1}{5}, \frac{2}{5})$

Finding the Slope of Line $D P$

We need the line joining $D (2, 2)$ and $P (- \frac{1}{5}, \frac{2}{5})$ .
Slope $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
$m = \frac{2 - \frac{2}{5}}{2 - ( - \frac{1}{5} )}$

Calculating the Slope

Numerator: $2 - \frac{2}{5} = \frac{8}{5}$
Denominator: $2 + \frac{1}{5} = \frac{11}{5}$
$m = \frac{8/5}{11/5} = \frac{8}{11}$

Equation of Line $D P$

Using point-slope form with $D (2, 2)$ : $y - y_{1} = m (x - x_{1})$
$y - 2 = \frac{8}{11} (x - 2)$
$11 (y - 2) = 8 (x - 2)$
$11 y - 22 = 8 x - 16$
$8 x - 11 y + 6 = 0$

Checking the Options

We need to find which given point lies on $8 x - 11 y + 6 = 0$ .
Let's test option $(- 9, - 6)$ :
$8 (- 9) - 11 (- 6) + 6 = - 72 + 66 + 6 = 0$
The point satisfies the equation!

00:00 / 00:00

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

Solution Diagram

Analyzing the Centroid

To find the centroid

D

of triangle

A B C

with vertices

A (3, - 1)

B (1, 3)

, and

C (2, 4)

, we use the centroid formula, which calculates the average of the coordinates:

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

Substituting the given values into the formula:

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

This simplifies to the anchor point $D (2, 2)$ .

Finding the Intersection Point

We now determine the intersection point

P

of the lines

x + 3 y - 1 = 0

and

3 x - y + 1 = 0

. We solve this system using the elimination method.

Multiplying the second equation by

3

yields:

9 x - 3 y = - 3

Adding this to the first equation,

x + 3 y = 1

, the

y

terms cancel out:

10 x = - 2 \Rightarrow x = - \frac{1}{5}

Substituting

x = - \frac{1}{5}

back into the first equation:

- \frac{1}{5} + 3 y = 1 \Rightarrow 3 y = \frac{6}{5} \Rightarrow y = \frac{2}{5}

Thus, our second anchor point is $P (- \frac{1}{5}, \frac{2}{5})$ .

Constructing the Line

We now find the equation of the line passing through

D (2, 2)

and

P (- \frac{1}{5}, \frac{2}{5})

. First, we calculate the slope

m

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}} = \frac{2 - \frac{2}{5}}{2 - ( - \frac{1}{5} )} = \frac{\frac{8}{5}}{\frac{11}{5}} = \frac{8}{11}

Using the point-slope form

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

with point

D (2, 2)

y - 2 = \frac{8}{11} (x - 2)

Cross-multiplying and simplifying:

11 (y - 2) = 8 (x - 2)

11 y - 22 = 8 x - 16

8 x - 11 y + 6 = 0

Final Verification

We test the point

(- 9, - 6)

against our derived line equation

8 x - 11 y + 6 = 0

8 (- 9) - 11 (- 6) + 6 = - 72 + 66 + 6 = 0

Since the result is zero, the point satisfies the equation. The line passing through the centroid and the intersection of the two given lines is $8 x - 11 y + 6 = 0$ .

Common Traps & FAQs

Why do we use the centroid formula? Can't we just find the intersection of medians?

You absolutely could, but that would be a much longer path! In competitive exams like JEE, efficiency is just as important as accuracy. The centroid formula

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

is a direct, elegant shortcut that saves you precious time.

Is there a faster way to find the intersection point

P

without solving the system?

While you might be tempted to look for a shortcut, solving the system of linear equations using elimination is the most robust and reliable method. It minimizes the chance of errors. Don't look for shortcuts that aren't there; trust the process.

How do I know which point lies on the line? Do I have to check all of them?

Yes, checking the options is a standard verification step. Once you have the equation of the line, simply substitute the

x

and

y

values of each option into the equation. The point that makes the equation equal to zero is your correct answer.

Animated Solution for Mathematics - Straight Lines: $A (3, - 1)$ , $B (1, 3)$ , $C (2, 4)$ are vertices of $Δ A B C$ if $D$ is centroid of $Δ A B C$ and $P$ is point of intersection of lines $x + 3 y - 1 = 0$ and $3 x - y + 1 = 0$ then which of the following points lies on line joining $D$ and $P$

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

The Centroid Formula

Substituting the Coordinates

Calculating Centroid $D$

Introducing the Intersecting Lines

Setting up Elimination

Solving for $x$

Solving for $y$

Finding the Slope of Line $D P$

Calculating the Slope

Equation of Line $D P$

Checking the Options

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

Analyzing the Centroid

Finding the Intersection Point

Constructing the Line

Final Verification

Similar Questions

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Select Answer:

Visualized Solution

Visualizing Triangle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC

The Centroid Formula

Substituting the Coordinates

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

Introducing the Intersecting Lines

Setting up Elimination

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

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

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

Calculating the Slope

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

Checking the Options

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

Analyzing the Centroid

Finding the Intersection Point

Constructing the Line

Final Verification

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Finding the Slope of Line $D P$

Equation of Line $D P$

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