JEE Main 2024 (08 Apr Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: The equations of two sides $A B$ and $A C$ of a triangle $A B C$ are $4 x + y = 14$ and $3 x - 2 y = 5$ , respectively. The point $(2, - 4/3)$ divides the third side $B C$ internally in the ratio $2 : 1$ . The equation of the side $B C$ is

Q: Why do we use parameterization instead of just finding the intersection of the lines?

Finding the intersection of $AB$ and $AC$ gives you vertex $A$, which is a great start, but it does not tell you where $B$ and $C$ are on those lines. Parameterization is the only way to 'lock' the points $B$ and $C$ onto their respective lines while allowing them to slide until they satisfy the condition at $P$. It is the most robust method for JEE problems involving points on lines.

Q: Is there a risk of making a sign error in the Section Formula?

Absolutely! The most common trap is mixing up the ratio $m:n$ with the coordinates. Always remember: the ratio $m$ multiplies the coordinate of the point $C$, and $n$ multiplies the point $B$. Keep your signs clean, and always double-check your arithmetic when dealing with negative fractions like $-\frac{4}{3}$.

Select Answer:

Progressive Hints

Hint 1

Visualize the triangle

A B C

. You are given the 'tracks' (the lines

A B

and

A C

) upon which points

B

and

C

must travel. Since you do not know the exact locations of

B

and

C

, you must define them using a single variable each. This is the art of parameterization—reducing the degrees of freedom to make the problem solvable.

Hint 2

The point

P (2, - \frac{4}{3})

is your anchor. It divides the segment

B C

in a

2 : 1

ratio. Recall the Section Formula: if a point divides a segment in ratio

m : n

, its coordinates are

\frac{m x _{2} + n x _{1}}{m + n}

and

\frac{m y _{2} + n y _{1}}{m + n}

. This is the bridge connecting your unknown variables to the known point

P

Hint 3

Start by writing

B

(x_{1}, 14 - 4 x_{1})

and

C

(x_{2}, \frac{3 x _{2} - 5}{2})

. Now, set up the two equations using the Section Formula for the

x

and

y

coordinates of

P

. You will have a system of two linear equations in

x_{1}

and

x_{2}

. Solve this system, and the rest is just finding the slope and applying the point-slope form.

Visualized Solution

Visualizing the Triangle Sides

Given side $A B$ : $4 x + y = 14$
Given side $A C$ : $3 x - 2 y = 5$
Point $P (2, - \frac{4}{3})$ lies on the third side $B C$ .

Parametrizing Point $B$

Since $B$ lies on $4 x + y = 14$ , let its x-coordinate be $x_{1}$ .
Then, $y_{1} = 14 - 4 x_{1}$ .
Coordinates of $B$ : $(x_{1}, 14 - 4 x_{1})$ .

Parametrizing Point $C$

Since $C$ lies on $3 x - 2 y = 5$ , let its x-coordinate be $x_{2}$ .
Then, $2 y = 3 x_{2} - 5 \Rightarrow y_{2} = \frac{3 x _{2} - 5}{2}$ .
Coordinates of $C$ : $(x_{2}, \frac{3 x _{2} - 5}{2})$ .

The Internal Division at $P$

Point $P (2, - \frac{4}{3})$ divides $B C$ internally in the ratio $2 : 1$ .
This means $B P : P C = 2 : 1$ .
Section Formula: $P = \frac{m C + n B}{m + n}$ where $m = 2, n = 1$ .

Applying Section Formula for $x$

For the x-coordinate of $P$ :
$2 = \frac{2 ( x _{2} ) + 1 ( x _{1} )}{2 + 1}$
$2 = \frac{2 x _{2} + x _{1}}{3}$
$2 x_{2} + x_{1} = 6 \dots (1)$

Applying Section Formula for $y$

For the y-coordinate of $P$ :
$- \frac{4}{3} = \frac{2 ( \frac{3 x _{2} - 5}{2} ) + 1 ( 14 - 4 x _{1} )}{2 + 1}$
Multiply by $3$ : $- 4 = 2 (\frac{3 x _{2} - 5}{2}) + (14 - 4 x_{1})$

Simplifying the $y$ Equation

Cancel the $2$ in the first term:
$- 4 = (3 x_{2} - 5) + (14 - 4 x_{1})$
$- 4 = 3 x_{2} - 4 x_{1} + 9$
$3 x_{2} - 4 x_{1} = - 13 \dots (2)$

Solving the System of Equations

From $(1)$ : $x_{1} = 6 - 2 x_{2}$
Substitute into $(2)$ :
$3 x_{2} - 4 (6 - 2 x_{2}) = - 13$
$3 x_{2} - 24 + 8 x_{2} = - 13$

Finding the Value of $x_{2}$

Combine like terms:
$11 x_{2} - 24 = - 13$
$11 x_{2} = 11$
$x_{2} = 1$

Finding the Value of $x_{1}$

Substitute $x_{2} = 1$ back into $x_{1} = 6 - 2 x_{2}$ :
$x_{1} = 6 - 2 (1)$
$x_{1} = 4$

Coordinates of $B$ and $C$

For $B$ : $x_{1} = 4 \Rightarrow y_{1} = 14 - 4 (4) = - 2 \Rightarrow B (4, - 2)$
For $C$ : $x_{2} = 1 \Rightarrow y_{2} = \frac{3 ( 1 ) - 5}{2} = - 1 \Rightarrow C (1, - 1)$

Finding the Slope of $B C$

Slope of $B C$ ( $m$ ) using $B (4, - 2)$ and $C (1, - 1)$ :
$m = \frac{- 1 - ( - 2 )}{1 - 4}$
$m = \frac{1}{- 3} = - \frac{1}{3}$

The Final Equation of Side $B C$

Equation of $B C$ using point $C (1, - 1)$ and slope $m = - \frac{1}{3}$ :
$y - (- 1) = - \frac{1}{3} (x - 1)$
$3 (y + 1) = - (x - 1)$
$3 y + 3 = - x + 1 \Rightarrow x + 3 y + 2 = 0$

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The Sigma Insight: Various Forms of Equations of a Line

Solution Diagram

Analyzing the Setup

We are given two lines,

A B

and

A C

, defined by the equations

4 x + y = 14

and

3 x - 2 y = 5

, respectively. We are also given a point

P (2, - \frac{4}{3})

that lies on the segment

B C

such that it divides

B C

in a ratio of

2 : 1

Our goal is to determine the coordinates of

B

and

C

to reconstruct the line

B C

The Art of Parameterization

To avoid an excess of variables, we parameterize the points

B

and

C

based on their respective lines. If

B

lies on

4 x + y = 14

, we can express its coordinates as:

B = (x_{1}, 14 - 4 x_{1})

Similarly, for point

C

on the line

3 x - 2 y = 5

, we solve for

y

to get

y = \frac{3 x - 5}{2}

. Thus, we define

C

as:

C = (x_{2}, \frac{3 x _{2} - 5}{2})

The Bridge of Section Formula

The point

P

divides the segment

B C

in the ratio

m : n = 2 : 1

. According to the Section Formula, the coordinates of

P

are given by:

P = (\frac{m x _{2} + n x _{1}}{m + n}, \frac{m y _{2} + n y _{1}}{m + n})

Substituting our values into this formula, we obtain the following system of equations:

2 = \frac{2 ( x _{2} ) + 1 ( x _{1} )}{3} \Rightarrow x_{1} + 2 x_{2} = 6

- \frac{4}{3} = \frac{2 ( \frac{3 x _{2} - 5}{2} ) + 1 ( 14 - 4 x _{1} )}{3}

The Algebraic Resolution

Simplifying the

y

-coordinate equation, the denominator

3

cancels out, leaving:

- 4 = 3 x_{2} - 5 + 14 - 4 x_{1}

- 4 = 3 x_{2} - 4 x_{1} + 9

4 x_{1} - 3 x_{2} = 13

We now solve the system: 1)

x_{1} = 6 - 2 x_{2}

4 x_{1} - 3 x_{2} = 13

Substituting (1) into (2):

4 (6 - 2 x_{2}) - 3 x_{2} = 13

24 - 8 x_{2} - 3 x_{2} = 13

- 11 x_{2} = - 11 \Rightarrow x_{2} = 1

Substituting

x_{2} = 1

back into (1), we find

x_{1} = 4

. Thus, the coordinates are $B (4, - 2)$ and $C (1, - 1)$ .

Final Calculation

To find the equation of line

B C

, we first calculate the slope

m

m = \frac{- 1 - ( - 2 )}{1 - 4} = \frac{1}{- 3} = - \frac{1}{3}

Using the point-slope form

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

with point

C (1, - 1)

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

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

3 y + 3 = - x + 1

The final equation of the line

B C

is $x + 3 y + 2 = 0$ .

Common Traps & FAQs

Why do we use parameterization instead of just finding the intersection of the lines?

Finding the intersection of

A B

and

A C

gives you vertex

A

, which is a great start, but it does not tell you where

B

and

C

are on those lines. Parameterization is the only way to 'lock' the points

B

and

C

onto their respective lines while allowing them to slide until they satisfy the condition at

P

. It is the most robust method for JEE problems involving points on lines.

Is there a risk of making a sign error in the Section Formula?

Absolutely! The most common trap is mixing up the ratio

m : n

with the coordinates. Always remember: the ratio

m

multiplies the coordinate of the point

C

, and

n

multiplies the point

B

. Keep your signs clean, and always double-check your arithmetic when dealing with negative fractions like

- \frac{4}{3}

What if the question asked for the length of

B C

instead of the equation?

The process remains identical! Once you have the coordinates of

B

and

C

(which we found as

(4, - 2)

and

(1, - 1)

), you would simply use the distance formula:

d = (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}

. The heavy lifting is always in finding the coordinates; the rest is just a quick calculation.

Animated Solution for Mathematics - Straight Lines: The equations of two sides $A B$ and $A C$ of a triangle $A B C$ are $4 x + y = 14$ and $3 x - 2 y = 5$ , respectively. The point $(2, - 4/3)$ divides the third side $B C$ internally in the ratio $2 : 1$ . The equation of the side $B C$ is

Select Answer:

Visualized Solution

Visualizing the Triangle Sides

Parametrizing Point $B$

Parametrizing Point $C$

The Internal Division at $P$

Applying Section Formula for $x$

Applying Section Formula for $y$

Simplifying the $y$ Equation

Solving the System of Equations

Finding the Value of $x_{2}$

Finding the Value of $x_{1}$

Coordinates of $B$ and $C$

Finding the Slope of $B C$

The Final Equation of Side $B C$

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Art of Parameterization

The Bridge of Section Formula

The Algebraic Resolution

Final Calculation

Similar Questions

In a $△ A B C$ , suppose $y = x$ is the equation of the bisector of the angle $B$ and the equation of the side $A C$ is $2 x - y = 2$ . If $2 A B = B C$ and the point $A$ and $B$ are respectively $(4, 6)$ and $(α, β)$ , then $α + 2 β$ is equal to

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

Select Answer:

Visualized Solution

Visualizing the Triangle Sides

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

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

The Internal Division at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P

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

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

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

Solving the System of Equations

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x2​

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x1​

Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

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

The Final Equation of Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } BC

The Sigma Insight: Various Forms of Equations of a Line

Analyzing the Setup

The Art of Parameterization

The Bridge of Section Formula

The Algebraic Resolution

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let PS be the median of the triangle with vertices P(2,2),Q(6,−1) and R(7,3). The equation of the line passing through (1,−1) and parallel to PS is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A straight line through the point A(3,4) is such that its intercept between the axes is bisected at A. Its equation is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let P=(−1,0),Q=(0,0) and R=(3,33​) be three points. The equation of the bisector of the angle PQR is

Parametrizing Point $B$

Parametrizing Point $C$

The Internal Division at $P$

Applying Section Formula for $x$

Applying Section Formula for $y$

Simplifying the $y$ Equation

Finding the Value of $x_{2}$

Finding the Value of $x_{1}$

Coordinates of $B$ and $C$

Finding the Slope of $B C$

The Final Equation of Side $B C$

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A straight line through the point $A (3, 4)$ is such that its intercept between the axes is bisected at $A$ . Its equation is

If in a parallelogram $A B D C$ , the coordinates of $A$ , $B$ and $C$ are respectively $(1, 2)$ , $(3, 4)$ and $(2, 5)$ then the equation of the diagonal $A D$ is:

One side of a rectangle lies along the line $4 x + 7 y + 5 = 0$ . Two of its vertices are $(- 3, 1)$ and $(1, 1)$ . Find the equations of the other three sides.

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