Q.Match the entries of Column $C_1$ with their appropriate answers given under Column $C_2$. Column $C_1$:

The core concept for parts (a) and (b) is the distance from a point to a line. For part (c), it is the section formula for trisection points.

(a) Points on $x+5y=13$ at a distance of 2 units from $12x-5y+26=0$:

1. Let a point on $x+5y=13$ be $(13-5y, y)$.
2. Using the distance formula with $A=12, B=-5, C=26$ and $d=2$: $\frac{|12(13-5y) - 5y + 26|}{\sqrt{12^2 + (-5)^2}} = 2$ $\frac{|156 - 60y - 5y + 26|}{\sqrt{144+25}} = 2$ $\frac{|182 - 65y|}{13} = 2 \implies |182 - 65y| = 26$.
3. Solving for $y$: $182 - 65y = 26 \implies 65y = 156 \implies y = \frac{12}{5}$. Then $x = 13 - 5(\frac{12}{5}) = 1$. Point: $\left(1, \frac{12}{5}\right)$. $182 - 65y = -26 \implies 65y = 208 \implies y = \frac{16}{5}$. Then $x = 13 - 5(\frac{16}{5}) = -3$. Point: $\left(-3, \frac{16}{5}\right)$. This matches option (iii).

(b) Points on $x+y=4$ at a unit distance from $4x+3y-10=0$:

1. Let a point on $x+y=4$ be $(x, 4-x)$.
2. Using the distance formula with $A=4, B=3, C=-10$ and $d=1$: $\frac{|4x + 3(4-x) - 10|}{\sqrt{4^2 + 3^2}} = 1$ $\frac{|4x + 12 - 3x - 10|}{\sqrt{16+9}} = 1$ $\frac{|x+2|}{5} = 1 \implies |x+2| = 5$.
3. Solving for $x$: $x+2 = 5 \implies x = 3$. Then $y = 4-3 = 1$. Point: $(3,1)$. $x+2 = -5 \implies x = -7$. Then $y = 4-(-7) = 11$. Point: $(-7,11)$. This matches option (i).

(c) Points of trisection of the line segment joining A$(-2,5)$ and B$(3,1)$: …

Let's break down each part of the problem. Parts (a) and (b) rely on the concept of the distance from a point to a line, while part (c) uses the section formula.

Concept: Distance From a Point to a Line

The perpendicular distance from a point $(x_0, y_0)$ to a line given by the equation $Ax+By+C=0$ is a fundamental concept in coordinate geometry. It represents the shortest distance between the point and any point on the line.

The absolute value in the numerator is crucial because distance is always non-negative. When we solve for coordinates, this absolute value will lead to two possible cases, corresponding to points on either side of the line (or in this problem, points on the given line that are at the specified distance from another line).

Concept: Section Formula

When a point divides a line segment joining two given points in a specific ratio, its coordinates can be found using the section formula.

Now, let's solve each part.

Part (a): Finding points on $x+5y=13$ at a distance of 2 units from $12x-5y+26=0$.

1. Represent a general point on the first line:

   We are looking for points on the line $x+5y=13$. To simplify calculations, we can express one coordinate in terms of the other. From $x+5y=13$, we get $x = 13-5y$. So, any point on this line can be represented as $(13-5y, y)$. Let's call this point $(x_0, y_0)$.

2. Apply the distance formula:

   The distance from $(x_0, y_0) = (13-5y, y)$ to the line $12x-5y+26=0$ must be 2 units. Using the distance formula:

We are given $d=2$, so:

3. Solve for $y$ using the absolute value: Multiplying by 13, we get:

This absolute value equation gives two possibilities:
*   **Case 1:** $182 - 65y = 26$
    $65y = 182 - 26$
    $65y = 156$
    $y = \frac{156}{65} = \frac{12 \times 13}{5 \times 13} = \frac{12}{5}$
*   **Case 2:** $182 - 65y = -26$
    $65y = 182 + 26$
    $65y = 208$
    $y = \frac{208}{65} = \frac{16 \times 13}{5 \times 13} = \frac{16}{5}$

4. Find the corresponding $x$ coordinates:

Using $x = 13-5y$:

* For $y = \frac{12}{5}$:

$x = 13 - 5\left(\frac{12}{5}\right) = 13 - 12 = 1$

This gives the point $\left(1, \frac{12}{5}\right)$.

* For $y = \frac{16}{5}$:

$x = 13 - 5\left(\frac{16}{5}\right) = 13 - 16 = -3$

This gives the point $\left(-3, \frac{16}{5}\right)$.

These are the two points P and Q.
Comparing with Column $C_2$, these points match option (iii).

Part (b): Finding points on $x+y=4$ at a unit distance from $4x+3y-10=0$.

1. Represent a general point on the first line:

   For the line $x+y=4$, we can write $x = 4-y$. So, any point on this line is $(4-y, y)$. Let this be $(x_0, y_0)$.

2. Apply the distance formula:

   The distance from $(x_0, y_0) = (4-y, y)$ to the line $4x+3y-10=0$ must be 1 unit.

Given $d=1$:

3. Solve for $y$ using the absolute value: Multiplying by 5:

Again, two possibilities:
*   **Case 1:** $6 - y = 5$
    $y = 6 - 5 = 1$
*   **Case 2:** $6 - y = -5$ …