JEE Main 2024 (31 Jan Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Straight Lines: Let $A (a, b), B (3, 4)$ and $(- 6, - 8)$ respectively denote the centroid, circumcentre and orthocentre of a triangle. Then, the distance of the point $P (2 a + 3, 7 b + 5)$ from the line $2 x + 3 y - 4 = 0$ measured parallel to the line $x - 2 y - 1 = 0$ is

Q: Wait, I got a negative value for $r$. Did I mess up the calculation?

Not at all! In the parametric form $x = x_1 + r \cos \theta$, the sign of $r$ simply indicates the direction relative to your chosen angle $\theta$. Since distance is a scalar magnitude, we take the absolute value. You are doing great!

Q: How do I remember the Euler line ratio? Is it $1:2$ or $2:1$?

Think of the alphabet: $H, G, O$. $H$ is the Orthocentre, $O$ is the Circumcentre. $G$ is the Centroid. The Centroid is 'central'—it is closer to the middle. It divides the segment $HO$ in the ratio $2:1$ starting from $H$. Just remember: $HG:GO = 2:1$.

Select Answer:

Progressive Hints

Hint 1

Before you dive into the coordinates, recall the hidden symmetry of a triangle. The orthocentre

H

, centroid

G

, and circumcentre

O

are not just random points; they are bound by the Euler line. Do you remember the specific ratio in which the centroid divides the segment connecting the orthocentre and the circumcentre?

Hint 2

Once you have the coordinates of

P

, you are asked for a distance 'measured parallel to' a line. This is a massive hint to avoid the standard perpendicular distance formula. Instead, think of the parametric form of a line:

x = x_{1} + r cos θ

and

y = y_{1} + r sin θ

. Can you find the slope of the line

x - 2 y - 1 = 0

to determine your

cos θ

and

sin θ

Hint 3

Now, set up your parametric equation. If

P

(3, 5)

and your direction is defined by

tan θ = \frac{1}{2}

, substitute

x = 3 + r cos θ

and

y = 5 + r sin θ

into the equation of the target line

2 x + 3 y - 4 = 0

. Solving for

r

will give you the directed distance. Remember to take the absolute value for the final answer!

Visualized Solution

The Euler Line Property

Orthocentre ( $H$ ), Centroid ( $G$ ), and Circumcentre ( $O$ ) are collinear.
The Centroid $G$ divides the segment $H O$ in the ratio $2 : 1$ internally.

Section Formula for Centroid

Using the section formula to find $G (a, b)$ .
$x = \frac{m x _{2} + n x _{1}}{m + n}$

Substituting Values for $a$

$a = \frac{2 ( 3 ) + 1 ( - 6 )}{2 + 1}$

Calculating $a$ and $b$

$a = \frac{6 - 6}{3} = 0$
$b = \frac{2 ( 4 ) + 1 ( - 8 )}{2 + 1} = 0$
So, $G (a, b) = (0, 0)$ .

Locating Point $P$

Point $P$ is given as $(2 a + 3, 7 b + 5)$ .
Substitute $a = 0, b = 0$ to get $P (3, 5)$ .

The Distance Problem

Find distance of $P (3, 5)$ from $L_{1} : 2 x + 3 y - 4 = 0$ .
The distance must be measured parallel to $L_{2} : x - 2 y - 1 = 0$ .

Direction of Measurement

The distance is measured along a line parallel to $L_{2}$ .
Slope of $L_{2}$ is $m = \frac{1}{2}$ .

Trigonometric Ratios

$tan θ = \frac{1}{2}$
$sin θ = \frac{1}{5}$
$cos θ = \frac{2}{5}$

Parametric Form of Line

Equation of line through $P (x_{1}, y_{1})$ at angle $θ$ :
$x = x_{1} + r cos θ$
$y = y_{1} + r sin θ$

Parametric Coordinates

Substitute $P (3, 5)$ and the trigonometric ratios:
$x = 3 + \frac{2 r}{5}$
$y = 5 + \frac{r}{5}$

Intersection with Target Line

This point must lie on $L_{1} : 2 x + 3 y - 4 = 0$ .
Substitute $x$ and $y$ :
$2 (3 + \frac{2 r}{5}) + 3 (5 + \frac{r}{5}) - 4 = 0$

Expanding the Equation

Expand the brackets:
$6 + \frac{4 r}{5} + 15 + \frac{3 r}{5} - 4 = 0$

Solving for $r$

Combine constants: $6 + 15 - 4 = 17$
Combine $r$ terms: $\frac{7 r}{5}$
$17 + \frac{7 r}{5} = 0$

Final Distance

$r = - \frac{17 5}{7}$
Distance is $∣ r ∣ = \frac{17 5}{7}$

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The Sigma Insight: Distance of a Point from a Line

Solution Diagram

Analyzing the Setup

The orthocentre

H (- 6, - 8)

and the circumcentre

O (3, 4)

define the Euler line of the triangle. The centroid

G (a, b)

lies on this line and divides the segment

H O

internally in the ratio

2 : 1

Using the section formula for a point dividing a line segment in the ratio

m : n

, the coordinates of

G

are given by:

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

Substituting the given values

H (- 6, - 8)

O (3, 4)

m = 2

, and

n = 1

G = (\frac{2 ( 3 ) + 1 ( - 6 )}{2 + 1}, \frac{2 ( 4 ) + 1 ( - 8 )}{2 + 1}) = (0, 0)

The Arithmetic Relief

With the centroid

G

identified as

(0, 0)

, we shift our focus to the point

P (3, 5)

. We are tasked with finding the distance from

P

to the line

L_{1} : 2 x + 3 y - 4 = 0

along a path parallel to the line

L_{2} : x - 2 y - 1 = 0

The slope of the line

L_{2}

m = \frac{1}{2}

. Since our path is parallel to

L_{2}

, the direction of our path is defined by

tan θ = \frac{1}{2}

From

tan θ = \frac{1}{2}

, we derive the trigonometric values:

cos θ = \frac{2}{5}, sin θ = \frac{1}{5}

The Parametric Magic

We express any point on the path passing through

P (3, 5)

in parametric form:

x = 3 + r cos θ = 3 + \frac{2 r}{5}

y = 5 + r sin θ = 5 + \frac{r}{5}

To find the distance

r

, we substitute these expressions into the equation of the target line

2 x + 3 y - 4 = 0

2 (3 + \frac{2 r}{5}) + 3 (5 + \frac{r}{5}) - 4 = 0

Expanding the equation:

6 + \frac{4 r}{5} + 15 + \frac{3 r}{5} - 4 = 0

17 + \frac{7 r}{5} = 0

Solving for

r

, we take the magnitude to represent the distance:

r = - \frac{17 5}{7} = \frac{17 5}{7}

The final distance is $\frac{17 5}{7}$ .

Common Traps & FAQs

Wait, I got a negative value for

r

. Did I mess up the calculation?

Not at all! In the parametric form

x = x_{1} + r cos θ

, the sign of

r

simply indicates the direction relative to your chosen angle

θ

. Since distance is a scalar magnitude, we take the absolute value. You are doing great!

Why can't I just use the perpendicular distance formula

\frac{∣ a x _{1} + b y _{1} + c ∣}{a ^{2} + b ^{2}}

That formula is specifically for the shortest distance, which is perpendicular. The problem explicitly asks for the distance measured *parallel* to another line. That is a different path entirely! Think of it as walking along a specific road rather than taking the shortest flight.

How do I remember the Euler line ratio? Is it

1 : 2

2 : 1

Think of the alphabet:

H, G, O

H

is the Orthocentre,

O

is the Circumcentre.

G

is the Centroid. The Centroid is 'central'—it is closer to the middle. It divides the segment

H O

in the ratio

2 : 1

starting from

H

. Just remember:

H G : GO = 2 : 1

Select Answer:

Visualized Solution

The Euler Line Property

Section Formula for Centroid

Substituting Values for $a$

Calculating $a$ and $b$

Locating Point $P$

The Distance Problem

Direction of Measurement

Trigonometric Ratios

Parametric Form of Line

Parametric Coordinates

Intersection with Target Line

Expanding the Equation

Solving for $r$

Final Distance

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Arithmetic Relief

The Parametric Magic

Similar Questions

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Select Answer:

Visualized Solution

The Euler Line Property

Section Formula for Centroid

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

Calculating .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

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

The Distance Problem

Direction of Measurement

Trigonometric Ratios

Parametric Form of Line

Parametric Coordinates

Intersection with Target Line

Expanding the Equation

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

Final Distance

The Sigma Insight: Distance of a Point from a Line

Analyzing the Setup

The Arithmetic Relief

The Parametric Magic

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 ?

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Substituting Values for $a$

Calculating $a$ and $b$

Locating Point $P$

Solving for $r$

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

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

Let $R$ be the point $(3, 7)$ and let $P$ and $Q$ be two points on the line $x + y = 5$ such that $P QR$ is an equilateral triangle. Then the area of $Δ P QR$ is :

Lines are drawn parallel to the line $4 x - 3 y + 2 = 0$ , at a distance $3/5$ from the origin. Then which one of the following points lies on any of these lines ?

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