# Write the equation of the family of lines with slope \frac{-1}{2} and find the two members of...

## Question:

Write the equation of the family of lines with slope {eq}\frac{-1}{2} {/eq} and find the two members of the family 5 units away from the origin.

## Family of lines.

Lines in a plane depend on two parameters: {eq}m {/eq} and {eq}q {/eq} if we consider the equation {eq}y=mx+q {/eq} of a line in the slope-intercept form. A family of lines is a set of lines depending on one parameter and there are two types of families of lines:

1. All lines in the plane with a fixed slope m.

2. All lines in the plane passing through a fixed point.

The distance between a point and a line is the shortest distance between the point and any point on the line. It is the length of the line segment that is perpendicular to the line and passes through the point.

The equation of the family of lines with fixed slope {eq}\displaystyle -\frac{1}{2} {/eq} is:

{eq}\displaystyle y= -\frac{1}{2}x +q {/eq}

with the parameter {eq}q \in \mathbb{R} {/eq}.

Lines in this family are all parallel.

To find a formula for the distance between the origin and a line in the family we first write the equation of the line that is perpendicular to every line of the family and passes through the origin:

{eq}y=2x {/eq}.

Then we find the coordinates of the intersection point between a line in the family and the line {eq}y=2x {/eq}. Substituting {eq}y=2x {/eq} in the equation of the family we obtain:

{eq}2x=-\frac{1}{2}x+q {/eq}

and solving for {eq}x {/eq} we find:

{eq}\displaystyle \frac{5}{2}x=q {/eq}

{eq}\displaystyle x=\frac{2}{5}q {/eq}.

The intersection point has coordinates:

{eq}\displaystyle (\frac{2}{5}q, \frac{4}{5}q) {/eq}.

Therefore the distance {eq}d {/eq} between the origin and a line in the family is given by :

{eq}d^2=\displaystyle \frac{4}{25}q^2 + \frac{16}{25}q^2 = \frac{20}{25}q^2= \frac{4}{5}q^2 {/eq}.

There are two members of the family which have distance equal to {eq}5 {/eq} from the origin, one with {eq}y {/eq}-intercept positive and one with {eq}y {/eq}-intercept negative and we find them solving for {eq}q {/eq} the quadratic equation {eq}d^2=5^2: {/eq}

{eq}\displaystyle \frac{4}{5}q^2 = 25 {/eq}

{eq}\displaystyle q^2 = 25\cdot \frac{5}{4} {/eq}.

The solutions are:

{eq}\displaystyle q_1 = 5\cdot \sqrt{\frac{5}{4}} =\frac{5}{2}\sqrt{5} {/eq} and

{eq}\displaystyle q_2 = -5\cdot \sqrt{\frac{5}{4}}= -\frac{5}{2}\sqrt{5} {/eq}.

Therefore the two members of the family 5 units away from the origin have equations:

{eq}\displaystyle y= -\frac{1}{2}x + \frac{5}{2}\sqrt{5} {/eq}

{eq}\displaystyle y= -\frac{1}{2}x - \frac{5}{2}\sqrt{5} {/eq}