Time for a few examples related to what I covered recently.

**Example 1 **Find the distance between the lines

(i) 4x + 3y = 5 and 4x + 3y = 10

(ii) y = x + 1 and 2x = 2y + 5

**Solution **The only thing what we need to do is transform both the given lines to the form ax + by + c = 0 (i.e. transform the equation so that RHS = 0).

Next, if both the lines have corresponding coefficients of x and y equal, we can apply the formula. Else, we’ll transform one of the equations by multiplying with some constant so that the coefficients become equal.

(i) The two lines can be written as 4x + 3y – 5 = 0 and 4x + 3y – 10 = 0. Therefore, the required distance is equal to |(-5)-(-10)|/\(\sqrt{4^2+3^2}\) = 1

(ii) The two equations can be written as x – y + 1 = 0 and 2x – 2y – 5 = 0. But before we can use the formula, we must multiply the first equation by 2, so that it becomes 2x – 2y + 2 = 0.

Now, the distance can be calculated as |2-(-5)|/\(\sqrt{2^2+(-2)^2}\) = 7/2\(\sqrt{2}\)

**Example 2 **The equations of two sides of a square are 3x + 4y – 5 = 0 and 3x + 4y – 15 = 0. If the third side passes through (6, 5), find the equations of the remaining sides.

**Solution **Note that the two given lines are parallel, and therefore are equations of opposite sides.

The other two sides will have equations of the form 4x – 3y + k = 0. (since they are perpendicular to the given sides).

Since one of them passes through (6,5) we get one value of k as -9, and therefore, one of the remaining sides as **4x – 3y – 9 = 0**.

What about the other one?

Since the equations represent sides of a square, the distance between the opposite sides will be equal.

That is, the distance between 3x + 4y – 5 = 0 and 3x + 4y – 15 = 0 will be same as the distance between 4x – 3y – 9 = 0 and the fourth side 4x – 3y + k = 0.

Hence we get the relation, |(-5)-(-15)|/\(\sqrt{3^2+4^2}\) = |-9-k|/\(\sqrt{4^2+(-3)^2}\), or |k+9|=10

This gives two values of k, 1 and -19, and therefore, two possible equations of the fourth side **4x – 3y + 1 = 0** and **4x – 3y – 19 = 0**.

**Example 3 **Find the equation of the lines parallel to the line 5x + 12y – 13 = 0, which are at a distance of 2 units from it.

**Solution **This one is easy. Any line parallel to the given line will be of the form 5x + 12y + k = 0. (explained here)

Now the distance between these two lines is |k+13|/\sqrt{5^2+12^2}\) which is given to be 2.

We get two values of k, 13 and -39, and two lines again: **5x + 12y + 13 = 0** and **5x + 12y – 39 = 0**. (lying on opposite sides of the given line.)