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Animated Solution for Mathematics - Straight Lines: If the line $2 x + y = k$ passes through the point which divides the line segment joining the points $(1, 1)$ and $(2, 4)$ in the ratio $3 : 2$ , then $k$ equals

Q: What if the ratio was external?

If the division were external, the formula would change to $C = (\frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n})$. Always check if the problem specifies internal or external division. In this case, it is internal.

Select Answer:

Visualized Solution

Visualizing Points $A$ and $B$

Given points: $A (1, 1)$ and $B (2, 4)$ .
Let's plot them and draw the line segment connecting them.

The Dividing Point $C$

A point $C (x, y)$ divides the segment $A B$ .
The ratio of division is $m : n = 3 : 2$ .

The Section Formula

To find $C (x, y)$ , we use the Section Formula for internal division.
$x = \frac{m x _{2} + n x _{1}}{m + n}$
$y = \frac{m y _{2} + n y _{1}}{m + n}$

Substituting the Values

Substitute $m = 3$ , $n = 2$ .
Point $A (x_{1}, y_{1}) = (1, 1)$ .
Point $B (x_{2}, y_{2}) = (2, 4)$ .
$x = \frac{3 ( 2 ) + 2 ( 1 )}{3 + 2}$
$y = \frac{3 ( 4 ) + 2 ( 1 )}{3 + 2}$

Computing the $x$ -coordinate

Simplify the numerator and denominator for $x$ .
$x = \frac{6 + 2}{5}$
$x = \frac{8}{5}$

Computing the $y$ -coordinate

Simplify the numerator and denominator for $y$ .
$y = \frac{12 + 2}{5}$
$y = \frac{14}{5}$
Point $C = (\frac{8}{5}, \frac{14}{5})$

Introducing the Line

The problem states a line passes through point $C$ .
The equation of the line is $2 x + y = k$ .

Applying the Point Condition

Since $C (\frac{8}{5}, \frac{14}{5})$ lies on the line, it must satisfy its equation.
Substitute $x = \frac{8}{5}$ and $y = \frac{14}{5}$ into $2 x + y = k$ .
$2 (\frac{8}{5}) + \frac{14}{5} = k$

Solving for $k$

Multiply and simplify the terms.
$\frac{16}{5} + \frac{14}{5} = k$
Combine the fractions: $\frac{16 + 14}{5} = k$

Final Calculation

$\frac{30}{5} = k$
$k = 6$
Final Answer: The value of $k$ is $6$ .

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The Sigma Insight: Distance and Section Formulas

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two points,

A (1, 1)

and

B (2, 4)

. These points are not just abstract numbers; they are anchors in space.

Our goal is to find a point

C

that sits on the segment connecting them, dividing it in a ratio of

3 : 2

. This is the essence of coordinate geometry: translating geometric relationships into algebraic precision.

The Section Formula

Our Bridge

To find the coordinates of

C

, we use the Section Formula. Think of this as a weighted average.

The

x

-coordinate and

y

-coordinate are given by the following expressions:

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

Here,

m = 3

and

n = 2

. Our points are

A (1, 1)

and

B (2, 4)

. Let us substitute these values carefully.

For the

x

-coordinate:

x = \frac{3 ( 2 ) + 2 ( 1 )}{3 + 2} = \frac{6 + 2}{5} = \frac{8}{5}

For the

y

-coordinate:

y = \frac{3 ( 4 ) + 2 ( 1 )}{3 + 2} = \frac{12 + 2}{5} = \frac{14}{5}

We have successfully pinpointed

C

(\frac{8}{5}, \frac{14}{5})

. This is a major milestone.

The Line Condition

The Final Lock

Now, we introduce the line

2 x + y = k

. The problem states that this line passes through

C

This is a powerful constraint. If a point lies on a line, its coordinates must satisfy the line's equation.

We substitute

x = \frac{8}{5}

and

y = \frac{14}{5}

into

2 x + y = k

2 (\frac{8}{5}) + \frac{14}{5} = k

Simplifying this, we get:

\frac{16}{5} + \frac{14}{5} = k

Adding the numerators, we find:

\frac{30}{5} = k

This simplifies beautifully to $k = 6$ . We have arrived at our answer.

Common Traps & FAQs

Why do we use the Section Formula instead of just finding the midpoint?

The midpoint formula is a special case of the Section Formula where the ratio is

1 : 1

. Here, the ratio is

3 : 2

, which means the point

C

is closer to

B

than

A

. The Section Formula is the generalized tool for any ratio

m : n

What if the ratio was external?

If the division were external, the formula would change to

C = (\frac{m x _{2} - n x _{1}}{m - n}, \frac{m y _{2} - n y _{1}}{m - n})

. Always check if the problem specifies internal or external division. In this case, it is internal.

Can I solve this using the slope of the line?

You could, but that would be unnecessarily complex. The most direct path is finding the point

C

first and then using the condition of point-on-line. Always look for the most efficient path in JEE problems.

Animated Solution for Mathematics - Straight Lines: If the line $2 x + y = k$ passes through the point which divides the line segment joining the points $(1, 1)$ and $(2, 4)$ in the ratio $3 : 2$ , then $k$ equals

Select Answer:

Visualized Solution

Visualizing Points $A$ and $B$

The Dividing Point $C$

The Section Formula

Substituting the Values

Computing the $x$ -coordinate

Computing the $y$ -coordinate

Introducing the Line

Applying the Point Condition

Solving for $k$

Final Calculation

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Section Formula

The Line Condition

Similar Questions

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Select Answer:

Visualized Solution

Visualizing Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } B

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

The Section Formula

Substituting the Values

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

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

Introducing the Line

Applying the Point Condition

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

Final Calculation

The Sigma Insight: Distance and Section Formulas

Analyzing the Setup

The Section Formula

The Line Condition

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 equation of the locus of a point equidistant from the point (a1​,b1​) and (a2​,b2​) is (a1​−a2​)x+(b1​−b2​)y+c=0, then the value of 'c' is

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Visualizing Points $A$ and $B$

The Dividing Point $C$

Computing the $x$ -coordinate

Computing the $y$ -coordinate

Solving for $k$

A straight line segment of length $l$ moves with its ends on two mutually perpendicular lines. Find the locus of the point which divides the line segment in the ratio $1 : 2$ .

If the equation of the locus of a point equidistant from the point $(a_{1}, b_{1})$ and $(a_{2}, b_{2})$ is $(a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0$ , then the value of 'c' is

For $a > b > c > 0$ , the distance between $(1, 1)$ and the point of intersection of the lines $a x + b y + c = 0$ and $b x + a y + c = 0$ is less than $22$ . Then

If $(1, 2), (4, 6), (5, 7)$ and $(a, b)$ are the vertices of a parallelogram $P QR S$ , then

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