Distance Formulas and Position of Points

Distance Formulas and Position of Points

Hello JEE aspirants! This lesson is crucial for mastering coordinate geometry, especially straight lines. Understanding distances and point positions is not just about memorizing formulas; it's about visualizing geometry algebraically. Expect direct questions and applications within more complex problems.

Distance of a Point from a Line

Imagine a point $P(x_1, y_1)$ and a line $ax + by + c = 0$. We want to find the shortest distance from the point to the line. Geometrically, this is the length of the perpendicular from $P$ to the line.

Explanation:

• $a, b, c$ are the coefficients from the line's equation.
• $x_1, y_1$ are the coordinates of the point.
• The absolute value ensures the distance is positive.
• $\sqrt{a^2 + b^2}$ normalizes the distance relative to the line's coefficients.

Intuition: The numerator $|ax_1 + by_1 + c|$ essentially 'plugs' the point into the line's equation. If the result is zero, the point lies on the line. The further away from zero, the greater the distance (before normalization).

Derivation (Conceptual): A more rigorous derivation involves vector projections, but for JEE, focus on application. Conceptually, the formula is derived by finding the length of the perpendicular dropped from the point onto the line. This involves finding the equation of the perpendicular line, determining the point of intersection, and then applying the standard distance formula between two points.

Example: Find the distance of the point $(2, 3)$ from the line $3x + 4y - 7 = 0$.

Solution: $$d = \frac{|3(2) + 4(3) - 7|}{\sqrt{3^2 + 4^2}} = \frac{|6 + 12 - 7|}{\sqrt{9 + 16}} = \frac{11}{5}$$

Distance Between Two Parallel Lines

Parallel lines have the same slope but different y-intercepts. Consider two parallel lines: $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$.

Explanation:

• $c_1$ and $c_2$ are the constant terms in the line equations.
• $a$ and $b$ are the coefficients of $x$ and $y$, which are the same for parallel lines.

Intuition: The distance is proportional to the difference in their constant terms, normalized by $\sqrt{a^2 + b^2}$.

Derivation (Conceptual): Imagine taking any point on one line and calculating its distance from the other line using the point-to-line distance formula. This simplifies to the above formula due to the lines being parallel, leading to a common term that cancels out.

Example: Find the distance between the lines $2x + y - 4 = 0$ and $2x + y + 1 = 0$.

Solution: $$d = \frac{|-4 - 1|}{\sqrt{2^2 + 1^2}} = \frac{5}{\sqrt{5}} = \sqrt{5}$$

Position of a Point Relative to a Line

A line divides the plane into two regions. The expression $ax_1 + by_1 + c$ can tell us which side of the line the point $(x_1, y_1)$ lies on.

• If $ax_1 + by_1 + c > 0$, the point lies on one side.
• If $ax_1 + by_1 + c < 0$, the point lies on the other side.
• If $ax_1 + by_1 + c = 0$, the point lies on the line.

Intuition: This relates to the sign of the result when substituting the point's coordinates into the line's equation. It directly indicates which "half-plane" the point occupies.

Important: This is only meaningful if you keep the line's equation in the form $ax + by + c = 0$. The sign depends on the arrangement of terms.

Example: Determine the position of points $(1, 1)$ and $(-1, -1)$ relative to the line $x + y - 2 = 0$.

For $(1, 1)$: $1 + 1 - 2 = 0$. The point lies on the line.

For $(-1, -1)$: $-1 + (-1) - 2 = -4 < 0$. The point lies on the side where the expression is negative.

Foot of Perpendicular from a Point to a Line

The foot of the perpendicular is the point where the perpendicular from a point to a line intersects the line. Let's say the foot of perpendicular from point $P(x_1, y_1)$ to the line $ax + by + c = 0$ is $(h, k)$.

Therefore, $x = x_1 - a \frac{ax_1 + by_1 + c}{a^2 + b^2}$ and $y = y_1 - b \frac{ax_1 + by_1 + c}{a^2 + b^2}$. These give you the coordinates of the foot of the perpendicular $(x, y)$.

Explanation:

• $(x_1, y_1)$ is the given point.
• $a, b, c$ are coefficients of the line.
• $(x, y)$ are the coordinates of the foot of the perpendicular.

Intuition: This formula cleverly uses the direction ratios of the normal to the line and scales them based on the perpendicular distance from the point to the line. The negative sign ensures we move in the correct direction toward the line.

Example: Find the foot of the perpendicular from $(1, 2)$ to the line $x + y - 1 = 0$.

Solution: $\frac{x - 1}{1} = \frac{y - 2}{1} = -\frac{(1 + 2 - 1)}{1^2 + 1^2} = -1$

So, $x = 1 - 1 = 0$ and $y = 2 - 1 = 1$. Foot of the perpendicular is $(0, 1)$.

Image of a Point About a Line

The image of a point is its reflection across the line. If $Q(x, y)$ is the image of point $P(x_1, y_1)$ about the line $ax + by + c = 0$, then:

Therefore, $x = x_1 - 2a \frac{ax_1 + by_1 + c}{a^2 + b^2}$ and $y = y_1 - 2b \frac{ax_1 + by_1 + c}{a^2 + b^2}$.

Explanation:

• $(x_1, y_1)$ is the original point.
• $a, b, c$ are coefficients of the line.
• $(x, y)$ are the coordinates of the image.

Intuition: It's similar to finding the foot of the perpendicular, but we go twice the distance along the normal from the point to reach its image.

Example: Find the image of the point $(1, 2)$ about the line $x + y - 1 = 0$.

Solution: $\frac{x - 1}{1} = \frac{y - 2}{1} = -\frac{2(1 + 2 - 1)}{1^2 + 1^2} = -2$

So, $x = 1 - 2 = -1$ and $y = 2 - 2 = 0$. The image is $(-1, 0)$.

Good luck mastering these concepts!