A straight line L through the origin...

MathematicsVarious Forms of Equations of a LineJEE Advanced 2002Moderate

A straight line $L$ through the origin meets the lines $x + y = 1$ and $x + y = 3$ at $P$ and $Q$ respectively. Through $P$ and $Q$ two straight lines $L_{1}$ and $L_{2}$ are drawn, parallel to $2 x - y = 5$ and $3 x + y = 5$ respectively. Lines $L_{1}$ and $L_{2}$ intersect at $R$ . Show that the locus of $R$ , as $L$ varies, is a straight line.

Visualized Solution (Hindi)

Setting the Stage

Given Lines: $x + y = 1$ and $x + y = 3$
Variable Line $L$ : Passes through $(0, 0)$ , let its equation be $y = m x$
Goal: Find the locus of the intersection point $R$ of lines $L_{1}$ and $L_{2}$

Defining Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L and Intersection Method

Let the equation of line $L$ be $y = m x$ .
Point $P$ is the intersection of $y = m x$ and $x + y = 1$ .
Point $Q$ is the intersection of $y = m x$ and $x + y = 3$ .

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

Substitute $y = m x$ into $x + y = 1$ : $x + m x = 1$
Factor out $x$ : $x (1 + m) = 1 \Rightarrow x_{P} = \frac{1}{1 + m}$
Find $y_{P}$ : $y_{P} = m \cdot x_{P} = \frac{m}{1 + m}$
Coordinates of $P$ : $(\frac{1}{1 + m}, \frac{m}{1 + m})$

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

Substitute $y = m x$ into $x + y = 3$ : $x + m x = 3$
Solve for $x$ : $x_{Q} = \frac{3}{1 + m}$
Solve for $y$ : $y_{Q} = \frac{3 m}{1 + m}$
Coordinates of $Q$ : $(\frac{3}{1 + m}, \frac{3 m}{1 + m})$

Logic Bridge: Defining Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{1} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{2}

Line $L_{1}$ is parallel to $2 x - y = 5$ , so its slope $m_{1} = 2$ .
Line $L_{2}$ is parallel to $3 x + y = 5$ , so its slope $m_{2} = - 3$ .
Point-Slope Form: $y - y_{1} = m (x - x_{1})$

Raw Setup: Equations of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{1} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L_{2} at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

For $R (X, Y)$ on $L_{1}$ : $Y - \frac{m}{1 + m} = 2 (X - \frac{1}{1 + m})$
Rearrange: $Y - 2 X = \frac{m - 2}{1 + m}$ ... (Eq. 1)
For $R (X, Y)$ on $L_{2}$ : $Y - \frac{3 m}{1 + m} = - 3 (X - \frac{3}{1 + m})$
Rearrange: $Y + 3 X = \frac{3 m + 9}{1 + m}$ ... (Eq. 2)

Atomic Compute: Isolating Terms with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

From Eq. 1: $Y - 2 X = \frac{m + 1 - 3}{1 + m} = 1 - \frac{3}{1 + m} \Rightarrow \frac{3}{1 + m} = 1 - Y + 2 X$
From Eq. 2: $Y + 3 X = \frac{3 ( m + 1 ) + 6}{1 + m} = 3 + \frac{6}{1 + m} \Rightarrow \frac{6}{1 + m} = Y + 3 X - 3$

Atomic Compute: Final Elimination

Substitute $\frac{6}{1 + m} = 2 \cdot (\frac{3}{1 + m})$ :
$Y + 3 X - 3 = 2 (1 - Y + 2 X)$
$Y + 3 X - 3 = 2 - 2 Y + 4 X$
Rearrange terms: $4 X - 3 X - 2 Y - Y + 2 + 3 = 0$
Final Locus: $X - 3 Y + 5 = 0$

The Way Forward

Key Takeaway: The locus of the intersection of lines passing through points on two parallel lines with fixed slopes is a straight line.
Next Challenge: What if the slopes of $L_{1}$ and $L_{2}$ were $m$ and $- m$ ? How would the locus change?

00:00 / 00:00

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Visualized Solution (Hindi)

Setting the Stage

Defining Line $L$ and Intersection Method

Raw Setup: Finding Point $P$

Raw Setup: Finding Point $Q$

Logic Bridge: Defining Slopes of $L_{1}$ and $L_{2}$

Raw Setup: Equations of $L_{1}$ and $L_{2}$ at $R$

Atomic Compute: Isolating Terms with $m$

Atomic Compute: Final Elimination

The Way Forward

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Visualized Solution (Hindi)

Setting the Stage

Defining Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L and Intersection Method

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

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

Logic Bridge: Defining Slopes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Atomic Compute: Isolating Terms with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Atomic Compute: Final Elimination

The Way Forward

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A rectangle PQRS has its side PQ parallel to the line y=mx and vertices P,Q and S on the lines y=a,x=b and x=−b, respectively. Find the locus of the vertex R.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P=(1,0),Q=(−1,0) and R=(2,0) are three given points, then locus of the point S satisfying the relation SQ2+SR2=2SP2, is

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A rectangle PQRS has its side PQ parallel to the line y=mx and vertices P,Q and S on the lines y=a,x=b and x=−b, respectively. Find the locus of the vertex R.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If P=(1,0),Q=(−1,0) and R=(2,0) are three given points, then locus of the point S satisfying the relation SQ2+SR2=2SP2, is

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

Defining Line $L$ and Intersection Method

Raw Setup: Finding Point $P$

Raw Setup: Finding Point $Q$

Logic Bridge: Defining Slopes of $L_{1}$ and $L_{2}$

Atomic Compute: Isolating Terms with $m$

A rectangle $P QR S$ has its side $P Q$ parallel to the line $y = m x$ and vertices $P, Q$ and $S$ on the lines $y = a, x = b$ and $x = - b$ , respectively. Find the locus of the vertex $R$ .

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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 $1 : 2$ .