Find the equation of the line passing through the point (2, 3) and making intercept of length 2 unit

Q: Why do we use the normal to the parallel lines to find the angle?

Great question! The distance between two parallel lines is defined as the shortest distance, which is measured along the common normal. By using the normal, we create a clear geometric relationship between the intercept, the distance, and the angle, turning a complex coordinate geometry problem into a simple right-triangle trigonometry problem.

Q: When solving the modulus equation, I got an impossible result like $4 = -1$. Did I make a mistake?

Not at all! That 'impossible' result is actually a mathematical signal. When the terms involving the slope $m$ cancel out in such a way that you are left with a contradiction, it means the slope $m$ is undefined, which corresponds to a vertical line. In this case, the vertical line $x = 2$ is indeed one of your solutions.

Q: Can we solve this using the distance of a point from a line formula instead?

Yes, you absolutely can! You could assume the line is $y - 3 = m(x - 2)$ and find the distance from the parallel lines to the points of intersection. However, that approach is algebraically much heavier. The geometric approach we used is more elegant and less prone to calculation errors.

Question

Find the equation of the line passing through the point $(2, 3)$ and making intercept of length 2 units between the lines $y + 2x = 3$ and $y + 2x = 5$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, looking at two parallel lines, $2x + y - 3 = 0$ and $2x + y - 5 = 0$. They stretch out infinitely, never touching, like two parallel tracks.

You are standing at the point $P(2, 3)$, and your mission is to draw a line through this point that bridges the gap between these tracks, cutting an intercept of exactly $2$ units.

### Phase 1: The Perpendicular Bridge

Before we can find the line, we must understand the gap. The shortest distance between our two parallel lines is the perpendicular distance $d$.

Using the formula $d = \frac{|c_1 - c_2|}{\sqrt{a^2 + b^2}}$, we plug in our values: $a = 2$, $b = 1$, $c_1 = -3$, and $c_2 = -5$.

$$d = \frac{|-3 - (-5)|}{\sqrt{2^2 + 1^2}} = \frac{2}{\sqrt{5}}$$

This $d$ is the foundation of our triangle.

### Phase 2: The Geometric Bridge

Now, imagine drawing a normal line perpendicular to our parallel lines. Our required line cuts across this normal at an angle $	heta$.

If we look at the intercept of length $L = 2$ as the hypotenuse of a right-angled triangle, the perpendicular distance $d$ becomes the adjacent side to the angle $	heta$.

$$\cos 	heta = \frac{d}{L} = \frac{2/\sqrt{5}}{2} = \frac{1}{\sqrt{5}}$$

With $\cos 	heta = \frac{1}{\sqrt{5}}$, we can easily find $	an 	heta$. Since $\sec 	heta = \sqrt{5}$, we use the identity $	an^2 	heta = \sec^2 	heta - 1$ to find $	an 	heta = \sqrt{5 - 1} = 2$.

### Phase 3: The Algebraic Dance

We know the slope of our parallel lines is $m_p = -2$. The normal, being perpendicular, must have a slope $m_n = \frac{-1}{-2} = \frac{1}{2}$.

We are looking for a line with slope $m$ that makes an angle $	heta$ with this normal. The formula for the angle between two lines is:

$$	an 	heta = \left| \frac{m - m_n}{1 + m \cdot m_n} ight|$$

Substituting our values, we get:

$$2 = \left| \frac{m - 1/2}{1 + m/2} ight| = \left| \frac{2m - 1}{2 + m} ight|$$

### Phase 4: The Final Reveal

This modulus equation gives us two paths. First, if we take the positive case, $2 = \frac{2m - 1}{2 + m}$, we get $4 + 2m = 2m - 1$, which leads to $4 = -1$.

This absurdity tells us the slope is infinite, meaning our line is vertical: **$x = 2$**.

Second, taking the negative case, $-2 = \frac{2m - 1}{2 + m}$, we get $-4 - 2m = 2m - 1$. This simplifies to $4m = -3$, or $m = -\frac{3}{4}$.

Using the point-slope form $y - 3 = -\frac{3}{4}(x - 2)$, we arrive at $4y - 12 = -3x + 6$, or **$3x + 4y - 18 = 0$**.

We have found our two lines: **$x - 2 = 0$** and **$3x + 4y - 18 = 0$**. Both perfectly bridge the gap.

Animated Solution for Mathematics - Straight Lines: Find the equation of the line passing through the point $(2, 3)$ and making intercept of length 2 units between the lines $y + 2 x = 3$ and $y + 2 x = 5$ .

Visualized Solution

The Setup: Parallel Lines and Point $P$

Distance Between Parallel Lines

Substituting Values for $d$

Calculating Distance $d$

Intercept Geometry

Relating Distance, Intercept, and Angle

Finding $cos θ$

Calculating $tan θ$

Slopes of the Lines

Angle Between Two Lines Formula

Solving for Slope $m$ (Case 1)

Solving for Slope $m$ (Case 2)

Final Equations of the Lines

The Sigma Insight: Angle Between Two Lines

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

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Visualized Solution

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

Distance Between Parallel Lines

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

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