A family of lines is a set of lines having one or two factors in common with each other. Straight lines can belong to two types of families: one where the slope is the same and one where the y-intercept is the same.

Consider the two straight lines

- a
_{1}x+b_{1}y+c_{1}=0 - a
_{2}x+b_{2}y+c_{2}=0

For any nonzero constant k, the equation of the form

- a
_{1}x+b_{1}y+c_{1}+k(a_{2}x+b_{2}y+c_{2})=0

being linear in x and y is an equation of a straight line.

If (x_{1},y_{1}) is the point of intersection of line 1 and 2, then it must satisfy both equations 1 and 2:

- a
_{1}x_{1}+b_{1}y_{1}+c_{1}=0 - a
_{1}x_{1}+b_{2}y_{1}+c_{2}=0

Next, we check whether the point (x_{1},y_{1}) lies on 3 or not. For this, we replace x by x_{1} and y by y_{1} in equation 3, and we have

- a
_{1}x_{1}+b_{1}y_{1}+c_{1}+k(a_{2}x_{1}+b_{2}y_{1}+c_{2})=0

Using equation 4 and 5 in equation 6, we get the following result:

0+k(0)=0

This shows that equation 6 is true for all k and for x=x_{1},y=y_{1}. Thus, the point (x_{1},y_{1}) lies on 6 for all k. Equation 6 represents the equation of the line through the point of intersection of lines 1 and 2. Since k is any real number, equation 6 shows that there will be an infinite number of lines through the point of intersection of lines 1 and 3.

We must have the particular value of k find the equation of a line, and this particular value of kcan be found with the help of some given conditions.

**Example 1: Find the equation of a line through the point (1,3) and the point of intersection of lines 2x−3y+4=0 and 4x+y−1=0.**

The equation of a straight line through the point of intersection of lines 2x−3y+4=0 and 4x+y−1=0 is given as

2x−3y+4+k(4x+y−1)=0

Since the required line passes through the point (1,3), this point must satisfy the equation, i.e.

2(1)−3(3)+4+k(4(1)+3−1)=0−3+6k=0

k=12

Putting this value of k in the required equation of a straight line, we have

2x−3y+4+12(4x+y−1)=0

4x−6y+8+4x+y−1=0

**Answer: 8x−5y+7=0**

**Example 2. Write the equation of the family of lines satisfying the following conditions:**

a) parallel to the x-axis

b) through the point (0, -1)

c) perpendicular to 2x+7y−9=0

**d) parallel to x+4y−12=0**

a) All lines parallel to the x-axis have a slope of zero; the y-intercept can be anything. So the family of lines is y=0x+b or just y=b.

b) All lines passing through the point (0, -1) have the same y-intercept, b=−1. The family of lines is y=mx−1.

c) First, we need to find the slope of the given line. Rewriting 2x+7y−9=0 in slope-intercept form, we get y=−27x+97. The slope of the line is −27, so we’re looking for the family of lines with slope 72.

The family of lines is y=72x+b.

d) Rewrite x+4y−12=0 in slope-intercept form: y=−14x+3. The slope is −14, so that’s also the slope of the family of lines we are looking for.

The family of lines is y=−14x+b.