JEE Main 2020 (9 January Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: If $C$ be the centroid of the triangle having vertices $(3, - 1), (1, 3)$ and $(2, 4)$ . Let $P$ be the point of intersection of the lines $x + 3 y - 1 = 0$ and $3 x - y + 1 = 0$ , then the line passing through the points $C$ and $P$ also passes through the point:

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. To find the centroid

C

of a triangle with vertices

A (x_{1}, y_{1})

B (x_{2}, y_{2})

, and

D (x_{3}, y_{3})

, you must calculate the average of the x-coordinates and the average of the y-coordinates separately. Think of it as finding the geometric balance point of the triangle.

Hint 2

Hint 2: The Arsenal. To find the intersection point

P

of two lines, you are solving a system of linear equations. Use the elimination method: multiply one or both equations by a constant so that the coefficients of

x

y

are additive inverses, then add the equations to eliminate one variable.

Hint 3

Hint 3: The Execution. Once you have points

C (x_{1}, y_{1})

and

P (x_{2}, y_{2})

, the slope

m

of the line passing through them is

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

. Use the point-slope form

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

to derive the equation of the line.

Visualized Solution

Visualizing the Triangle

Given vertices of the triangle: $A (3, - 1)$ , $B (1, 3)$ , and $D (2, 4)$ .
We need to find the centroid $C$ of this triangle.

Centroid Formula

The centroid $C (x, y)$ of a triangle with vertices $(x_{1}, y_{1})$ , $(x_{2}, y_{2})$ , $(x_{3}, y_{3})$ is given by:
$x = \frac{x _{1} + x _{2} + x _{3}}{3}$
$y = \frac{y _{1} + y _{2} + y _{3}}{3}$

Substituting the Vertices

Substituting the coordinates of $A (3, - 1)$ , $B (1, 3)$ , and $D (2, 4)$ :
$x = \frac{3 + 1 + 2}{3}$
$y = \frac{- 1 + 3 + 4}{3}$

Calculating Centroid $C$

$x = \frac{6}{3} = 2$
$y = \frac{6}{3} = 2$
Therefore, the centroid is $C (2, 2)$ .

Introducing the Intersecting Lines

We are given two lines:
Line 1: $x + 3 y - 1 = 0 ⟹ x + 3 y = 1$
Line 2: $3 x - y + 1 = 0 ⟹ 3 x - y = - 1$
Let $P$ be their point of intersection.

Solving for Intersection Point $P$

To find $P$ , we solve the system of linear equations:
(i) $x + 3 y = 1$
(ii) $3 x - y = - 1$
Multiply equation (ii) by $3$ to eliminate $y$ .

Eliminating $y$ to find $x$

Multiply (ii) by 3: $9 x - 3 y = - 3$ ... (iii)
Add (i) and (iii):
$(x + 3 y) + (9 x - 3 y) = 1 + (- 3)$
$10 x = - 2 ⟹ x = - \frac{1}{5}$

Finding the $y$ -coordinate of $P$

Substitute $x = - \frac{1}{5}$ into equation (ii): $3 x - y = - 1$
$3 (- \frac{1}{5}) - y = - 1$
$- \frac{3}{5} - y = - 1 ⟹ y = 1 - \frac{3}{5} = \frac{2}{5}$
Point $P$ is $(- \frac{1}{5}, \frac{2}{5})$ .

Slope of Line $C P$

We need the equation of the line passing through $C (2, 2)$ and $P (- \frac{1}{5}, \frac{2}{5})$ .
First, find the slope $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$ .

Calculating the Slope $m$

Substitute $C (2, 2)$ and $P (- \frac{1}{5}, \frac{2}{5})$ :
$m = \frac{2 - \frac{2}{5}}{2 - ( - \frac{1}{5} )}$
$m = \frac{\frac{10 - 2}{5}}{\frac{10 + 1}{5}} = \frac{\frac{8}{5}}{\frac{11}{5}}$
$m = \frac{8}{11}$

Equation of Line $C P$

Using the point-slope form: $y - y_{1} = m (x - x_{1})$
Using point $C (2, 2)$ and slope $m = \frac{8}{11}$ :
$y - 2 = \frac{8}{11} (x - 2)$

Simplifying the Line Equation

Cross-multiply by 11:
$11 (y - 2) = 8 (x - 2)$
$11 y - 22 = 8 x - 16$
Rearranging terms: $8 x - 11 y + 6 = 0$

Checking the Options

We need to find which given point lies on $8 x - 11 y + 6 = 0$ .
Let's test option (2): $(- 9, - 6)$
Substitute $x = - 9, y = - 6$ :
$8 (- 9) - 11 (- 6) + 6 = - 72 + 66 + 6 = 0$
The equation is satisfied.

Final Conclusion

The point $(- 9, - 6)$ satisfies the equation of the line $C P$ .
Therefore, the line passing through $C$ and $P$ also passes through $(- 9, - 6)$ .
Correct Option: (2)

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The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Centroid

To find the centroid

C

of the triangle with vertices

A (3, - 1)

B (1, 3)

, and

D (2, 4)

, we calculate the average of the coordinates. The centroid acts as the geometric center of the triangle.

The coordinates are determined by the following formulas:

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

Substituting the given values:

x = \frac{3 + 1 + 2}{3} = 2, y = \frac{- 1 + 3 + 4}{3} = 2

Thus, the centroid is located at

C (2, 2)

The Intersection of Paths

We now determine the intersection point

P

of the two lines defined by

x + 3 y - 1 = 0

and

3 x - y + 1 = 0

. We solve the system of equations:

x + 3 y = 1

3 x - y = - 1

To eliminate

y

, we multiply the second equation by

3

9 x - 3 y = - 3

Adding this to the first equation yields:

(x + 3 y) + (9 x - 3 y) = 1 - 3

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

Substituting

x = - \frac{1}{5}

into the first equation:

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

The intersection point is

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

Determining the Line Equation

We now find the equation of the line passing through

C (2, 2)

and

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

. First, we calculate the slope

m

m = \frac{\frac{2}{5} - 2}{- \frac{1}{5} - 2} = \frac{- \frac{8}{5}}{- \frac{11}{5}} = \frac{8}{11}

Using the point-slope form

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

with point

C (2, 2)

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

Multiplying by

11

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 equation

8 x - 11 y + 6 = 0

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

Since the equation is satisfied, the line passing through the centroid and the intersection point is indeed $8 x - 11 y + 6 = 0$ .

Common Traps & FAQs

Wait, why do we use the average of the vertices for the centroid? Is there a deeper reason?

That is a fantastic question! The centroid is the geometric center of the triangle, which is also its center of mass if the triangle were made of a uniform material. Mathematically, it is the point where the three medians intersect. The formula

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

is derived from the fact that the centroid divides each median in a

2 : 1

ratio. It is the most elegant way to find the balance point!

Why did we choose the elimination method for the intersection point

P

instead of substitution?

Both methods are perfectly valid! However, in this specific case, the coefficients of

y

in the equations

x + 3 y = 1

and

3 x - y = - 1

are

3

and

- 1

. By multiplying the second equation by

3

, we get

- 3 y

, which perfectly cancels out the

+ 3 y

in the first equation. It is often faster and less prone to arithmetic errors than substitution when dealing with fractions.

How do I know which point lies on the line without checking every single option?

Great intuition! You don't necessarily have to check every option if you are confident in your line equation. Once you have the equation

8 x - 11 y + 6 = 0

, you simply substitute the coordinates

(x, y)

of each option into the equation. If the result is

0

, that point lies on the line. It is a quick and reliable verification step.

Select Answer:

Visualized Solution

Visualizing the Triangle

Centroid Formula

Substituting the Vertices

Calculating Centroid $C$

Introducing the Intersecting Lines

Solving for Intersection Point $P$

Eliminating $y$ to find $x$

Finding the $y$ -coordinate of $P$

Slope of Line $C P$

Calculating the Slope $m$

Equation of Line $C P$

Simplifying the Line Equation

Checking the Options

Final Conclusion

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Centroid

The Intersection of Paths

Determining the Line Equation

Final Verification

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Triangle

Centroid Formula

Substituting the Vertices

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

Introducing the Intersecting Lines

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

Eliminating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y to find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Finding 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; } P

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

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

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

Simplifying the Line Equation

Checking the Options

Final Conclusion

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Centroid

The Intersection of Paths

Determining the Line Equation

Final Verification

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Let $P = (- 1, 0), Q = (0, 0)$ and $R = (3, 33)$ be three points. Then the equation of the bisector of the angle $P QR$ is

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