The distance of the point (2, 3) from the line 2x - 3y + 28 = 0, measured parallel to the line √3x -

Q: Wait, why can't we just use the perpendicular distance formula $\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}$?

That is a fantastic question! The perpendicular distance formula only gives you the *shortest* distance. Here, the problem explicitly constrains your movement to a specific direction (parallel to another line). You are essentially walking along a slanted path until you hit the target line, which is a different geometric scenario than the shortest perpendicular drop.

Q: How do I know the angle $\theta$ is $60^\circ$?

Look at the direction line: $\sqrt{3}x - y + 1 = 0$. If you rewrite this in the slope-intercept form $y = mx + c$, you get $y = \sqrt{3}x + 1$. The slope $m$ is $\sqrt{3}$. Since the slope of a line is defined as $\tan \theta$, we have $\tan \theta = \sqrt{3}$. The angle whose tangent is $\sqrt{3}$ is $60^\circ$.

Q: What if the value of $r$ comes out negative? Do I reject it?

In this context, $r$ represents the displacement along the line. If you get a negative value, it simply means the target line lies in the opposite direction of the vector defined by your angle. However, since the question asks for 'distance', we take the absolute magnitude. In this specific problem, the algebra will naturally lead you to the correct positive distance.

Question

The distance of the point $(2, 3)$ from the line $2x - 3y + 28 = 0$, measured parallel to the line $\sqrt{3}x - y + 1 = 0$, is equal to

Options:

(A) $4\sqrt{2}$

(B) $6\sqrt{3}$

(C) $3 + 4\sqrt{2}$

(D) $4 + 6\sqrt{3}$

Sigma Percentile · Accepted Answer

### The Geometry of the Path

Imagine you are standing at the point $A(2, 3)$ on a vast coordinate plane. You have a destination—a target line defined by the equation $2x - 3y + 28 = 0$.

If you were a bird, you would fly the shortest path, dropping perpendicularly onto the line. But you are not a bird; you are a traveler constrained to a specific road. You must walk parallel to the line $\sqrt{3}x - y + 1 = 0$.

This constraint changes everything. It turns a simple distance problem into a beautiful exercise in parametric geometry.

### Phase 1: Finding Your Compass

Before we take a single step, we must know our heading. The line that dictates our direction is $\sqrt{3}x - y + 1 = 0$.

To understand its tilt, we rearrange it into the slope-intercept form, $y = mx + c$. This gives us $y = \sqrt{3}x + 1$. The slope $m$ is $\sqrt{3}$.

We know that the slope of a line is the tangent of the angle $	heta$ it makes with the positive x-axis. So, $	an 	heta = \sqrt{3}$. Recalling our trigonometry, we identify that $	heta = 60^\circ$. Our path is set; we are walking at a $60^\circ$ angle relative to the horizontal.

### Phase 2: The Parametric Bridge

Now, how do we mathematically describe our journey? We use the parametric form of a straight line. If we start at $(x_1, y_1)$ and walk a distance $r$ at an angle $	heta$, any point $P(x, y)$ on our path is given by the elegant equations:

$$x = x_1 + r \cos 	heta$$

$$y = y_1 + r \sin 	heta$$

Substituting our starting point $A(2, 3)$ and our angle $	heta = 60^\circ$, we get:

$$x = 2 + r \cos 60^\circ = 2 + \frac{r}{2}$$

$$y = 3 + r \sin 60^\circ = 3 + \frac{\sqrt{3}r}{2}$$

This is the bridge between our movement and the target line. Every point $P$ on our path is now defined by the distance $r$ we have traveled.

### Phase 3: The Intersection

Our destination is the target line $2x - 3y + 28 = 0$. Since our point $P$ must lie on this line, its coordinates must satisfy the equation.

We substitute our parametric expressions for $x$ and $y$ into the line equation:

$$2\left(2 + \frac{r}{2}ight) - 3\left(3 + \frac{\sqrt{3}r}{2}ight) + 28 = 0$$

Now, we expand carefully. Precision is key here. We get:

$$4 + r - 9 - \frac{3\sqrt{3}r}{2} + 28 = 0$$

Grouping the constants and the $r$-terms, we have:

$$(4 - 9 + 28) + r\left(1 - \frac{3\sqrt{3}}{2}ight) = 0$$

$$23 + r\left(\frac{2 - 3\sqrt{3}}{2}ight) = 0$$

### Phase 4: The Final Calculation

We are almost there. Isolating $r$, we find:

$$r\left(\frac{2 - 3\sqrt{3}}{2}ight) = -23$$

$$r = \frac{-46}{2 - 3\sqrt{3}} = \frac{46}{3\sqrt{3} - 2}$$

To finish, we rationalize the denominator by multiplying the numerator and denominator by the conjugate, $3\sqrt{3} + 2$:

$$r = \frac{46(3\sqrt{3} + 2)}{(3\sqrt{3})^2 - 2^2} = \frac{46(3\sqrt{3} + 2)}{27 - 4} = \frac{46(3\sqrt{3} + 2)}{23}$$

Watch the magic happen: $46$ divided by $23$ is exactly $2$. Thus, $r = 2(3\sqrt{3} + 2) = 6\sqrt{3} + 4$.

We have arrived. The distance is **$4 + 6\sqrt{3}$**. It is a beautiful result, born from the simple act of following a path.

Animated Solution for Mathematics - Straight Lines: The distance of the point $(2, 3)$ from the line $2 x - 3 y + 28 = 0$ , measured parallel to the line $3 x - y + 1 = 0$ , is equal to

Select Answer:

Visualized Solution

Visualizing the Problem

Finding the Direction Angle $θ$

The Parametric Tool

Raw Setup for Point $P$

Substitution into Target Line

Expanding the Equation

Grouping and Simplifying

Isolating $r$

Rationalizing the Denominator

The Final Answer

The Sigma Insight: Various Forms of Equations of a Line

The Geometry of the Path

Phase 1

Phase 2

Phase 3

Phase 4

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