JEE Main 2024 (29 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Let $A$ be the point of intersection of the lines $3 x + 2 y = 14$ , $5 x - y = 6$ and $B$ be the point of intersection of the lines $4 x + 3 y = 8$ , $6 x + y = 5$ . The distance of the point $P (5, - 2)$ from the line $A B$ is

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Visualized Solution

Visualizing the Problem

Given point $P (5, - 2)$ .
Objective: Find the perpendicular distance from $P$ to line $A B$ .

Finding Point $A$ : The Setup

Point $A$ is the intersection of:
1) $3 x + 2 y = 14$
2) $5 x - y = 6$

Finding Point $A$ : Substitution

From (2): $y = 5 x - 6$
Substitute into (1): $3 x + 2 (5 x - 6) = 14$

Finding Point $A$ : Solving

$3 x + 10 x - 12 = 14$
$13 x = 26 ⟹ x = 2$
$y = 5 (2) - 6 = 4$
Point $A = (2, 4)$

Finding Point $B$ : The Setup

Point $B$ is the intersection of:
3) $4 x + 3 y = 8$
4) $6 x + y = 5$

Finding Point $B$ : Solving

From (4): $y = 5 - 6 x$
Substitute into (3): $4 x + 3 (5 - 6 x) = 8$
$- 14 x = - 7 ⟹ x = \frac{1}{2}$
$y = 5 - 6 (\frac{1}{2}) = 2$
Point $B = (\frac{1}{2}, 2)$

Equation of Line $A B$ : Slope

Slope $m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
$m = \frac{4 - 2}{2 - \frac{1}{2}}$
$m = \frac{2}{\frac{3}{2}} = \frac{4}{3}$

Equation of Line $A B$ : Point-Slope Form

Using point $A (2, 4)$ and $m = \frac{4}{3}$ :
$y - y_{1} = m (x - x_{1})$
$y - 4 = \frac{4}{3} (x - 2)$

General Equation of Line $A B$

$3 (y - 4) = 4 (x - 2)$
$3 y - 12 = 4 x - 8$
$4 x - 3 y + 4 = 0$

The Distance Formula

Distance $d = \frac{∣ A x _{1} + B y _{1} + C ∣}{A ^{2} + B ^{2}}$
Line: $4 x - 3 y + 4 = 0$
Point: $P (5, - 2)$

Distance Calculation: Substitution

Substitute $A = 4, B = - 3, C = 4$
Substitute $x_{1} = 5, y_{1} = - 2$
$d = \frac{∣4 ( 5 ) - 3 ( - 2 ) + 4∣}{4 ^{2} + ( - 3 ) ^{2}}$

Final Calculation

$d = \frac{∣20 + 6 + 4∣}{16 + 9}$
$d = \frac{30}{25}$
$d = \frac{30}{5} = 6$

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The Sigma Insight: Distance of a Point from a Line

Solution Diagram

Analyzing the Setup

We are tasked with finding the perpendicular distance from a point

P (5, - 2)

to a line

A B

. This line

A B

is defined by the intersection points of two pairs of linear equations.

Finding the Islands

Points A and B

Point

A

is the intersection of the lines

3 x + 2 y = 14

and

5 x - y = 6

. By isolating

y

in the second equation, we obtain

y = 5 x - 6

Substituting this into the first equation:

3 x + 2 (5 x - 6) = 14

Simplifying the expression:

3 x + 10 x - 12 = 14

13 x = 26

x = 2, y = 4

Thus, point

A

(2, 4)

For point

B

, we solve the system

4 x + 3 y = 8

and

6 x + y = 5

. From the second equation,

y = 5 - 6 x

Substituting into the first equation:

4 x + 3 (5 - 6 x) = 8

4 x + 15 - 18 x = 8

- 14 x = - 7

x = \frac{1}{2}, y = 2

Thus, point

B

(\frac{1}{2}, 2)

Building the Bridge

The Equation of Line AB

Now that we have points

A (2, 4)

and

B (\frac{1}{2}, 2)

, we calculate the slope

m

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

Using the point-slope form

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

with point

A (2, 4)

y - 4 = \frac{4}{3} (x - 2)

Rearranging into the general form

A x + B y + C = 0

3 (y - 4) = 4 (x - 2)

3 y - 12 = 4 x - 8

4 x - 3 y + 4 = 0

The Final Leap

Calculating Distance

We calculate the perpendicular distance

d

from

P (5, - 2)

to the line

4 x - 3 y + 4 = 0

using the formula:

d = \frac{∣ A x _{0} + B y _{0} + C ∣}{A ^{2} + B ^{2}}

Substituting

A = 4

B = - 3

C = 4

x_{0} = 5

, and

y_{0} = - 2

d = \frac{∣4 ( 5 ) - 3 ( - 2 ) + 4∣}{4 ^{2} + ( - 3 ) ^{2}}

d = \frac{∣20 + 6 + 4∣}{16 + 9}

d = \frac{30}{25} = \frac{30}{5} = 6

The perpendicular distance is 6 units.

Common Traps & FAQs

Wait, why do we need to find the equation of the line

A B

? Can't we just find the distance from

P

A

and

P

B

That is a classic trap! The distance from a point to a line is the shortest perpendicular distance, not the distance to the endpoints of a segment. If you calculate the distance to

A

B

, you are finding the length of a line segment, not the perpendicular distance to the infinite line passing through them.

Is there a way to find the equation of line

A B

without solving for the intersection points

A

and

B

Yes, there is! You could use the concept of the 'family of lines' passing through the intersection of two lines, but for JEE Advanced, solving for the points is often safer and less prone to conceptual errors unless the intersection points are messy fractions. Stick to the method that gives you the most clarity.

Why do we use the absolute value in the distance formula? Can distance be negative?

Distance is a scalar quantity and must always be positive. The expression

A x_{0} + B y_{0} + C

can result in a negative value depending on which side of the line the point

P

lies. The absolute value bars ensure that we are measuring the magnitude of the separation, keeping our physics and geometry consistent.

Animated Solution for Mathematics - Straight Lines: Let $A$ be the point of intersection of the lines $3 x + 2 y = 14$ , $5 x - y = 6$ and $B$ be the point of intersection of the lines $4 x + 3 y = 8$ , $6 x + y = 5$ . The distance of the point $P (5, - 2)$ from the line $A B$ is

Select Answer:

Visualized Solution

Visualizing the Problem

Finding Point $A$ : The Setup

Finding Point $A$ : Substitution

Finding Point $A$ : Solving

Finding Point $B$ : The Setup

Finding Point $B$ : Solving

Equation of Line $A B$ : Slope

Equation of Line $A B$ : Point-Slope Form

General Equation of Line $A B$

The Distance Formula

Distance Calculation: Substitution

Final Calculation

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

Finding the Islands

Building the Bridge

The Final Leap

Similar Questions

Let the point $P (α, β)$ be at a unit distance from each of the two lines $L_{1} : 3 x - 4 y + 12 = 0$ , and $L_{2} : 8 x + 6 y + 11 = 0$ . If $P$ lies below $L_{1}$ and above $L_{2}$ , then $100 (α + β)$ is equal to

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Select Answer:

Visualized Solution

Visualizing the Problem

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

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

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

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

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

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

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

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

The Distance Formula

Distance Calculation: Substitution

Final Calculation

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

Finding the Islands

Building the Bridge

The Final Leap

Similar Questions

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Lines are drawn parallel to the line 4x−3y+2=0, at a distance 3/5 from the origin. Then which one of the following points lies on any of these lines ?

Finding Point $A$ : The Setup

Finding Point $A$ : Substitution

Finding Point $A$ : Solving

Finding Point $B$ : The Setup

Finding Point $B$ : Solving

Equation of Line $A B$ : Slope

Equation of Line $A B$ : Point-Slope Form

General Equation of Line $A B$

If the line, $2 x - y + 3 = 0$ is at a distance $\frac{1}{5}$ and $\frac{2}{5}$ from the lines $4 x - 2 y + α = 0$ and $6 x - 3 y + β = 0$ , respectively, then the sum of all possible values of $α$ and $β$ is

The line $L$ given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $(13, 32)$ . The line $K$ is parallel to $L$ and has the equation $\frac{x}{c} + \frac{y}{3} = 1$ . Then the distance between $L$ and $K$ is

Let the equation $x (x + 2) (12 - k) = 2$ have equal roots. Then the distance of the point $(k, \frac{k}{2})$ from the line $3 x + 4 y + 5 = 0$ is

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