# Linear Function

**Linear function** is given as $$${y}={f{{\left({x}\right)}}}={m}{x}+{b}$$$.

$$${m}$$$ is called **slope** and $$${b}$$$ is called **y-intercept.** Graph of the linear function is **line.** Since there are two parameters in the linear function (m and b) it is enough two points to uniquely identify line.

If line passes through 2 points $$${\left({x}_{{1}},{y}_{{1}}\right)}$$$ and $$${\left({x}_{{2}},{y}_{{2}}\right)}$$$ then $$${\color{blue}{{{m}=\frac{{{y}_{{2}}-{y}_{{1}}}}{{{x}_{{2}}-{x}_{{1}}}}}}}$$$.

$$${m}$$$ shows rate of change of function. For example, for function $$${y}={3}{x}+{2}$$$ if we increase $$${x}$$$ by 1 unit, then $$${y}$$$ will increase by 3 units. A characteristic feature of linear functions is that they grow at a constant rate.

Sometimes linear function can be given in the form $$${a}{x}+{c}{y}={d}$$$, where $$${c}\ne{0}$$$. We can easily convert it into standard form by dividing both sides of equation by $$${c}$$$: $$${y}=-\frac{{a}}{{c}}{x}+\frac{{d}}{{c}}$$$. Here $$${m}=-\frac{{a}}{{c}}$$$ and $$${b}=\frac{{d}}{{c}}$$$.

**Example 1.** Find equation of line that passes through points $$${\left({2},-{3}\right)}$$$ and $$${\left(-{5},{1}\right)}$$$.

Slope of the line is $$${m}=\frac{{{1}-{\left(-{3}\right)}}}{{-{5}-{2}}}=-\frac{{4}}{{7}}$$$.

Now, equation of line can be rewritten as $$${y}=-\frac{{4}}{{7}}{x}+{b}$$$. Since line passes through point $$${\left({2},-{3}\right)}$$$ then $$$-{3}=-\frac{{4}}{{7}}\cdot{2}+{b}$$$. From this $$${b}=-\frac{{13}}{{7}}$$$.

Thus, equation of line is $$${y}=-\frac{{4}}{{7}}{x}-\frac{{13}}{{7}}$$$.

**Example 2.** Draw line whose equation is $$${3}{x}+{2}{y}={6}$$$.

We need to points to draw a graph.

If $$${x}={0}$$$ then $$${3}\cdot{0}+{2}{y}={6}$$$ or $$${y}={3}$$$. Therefore, first point is $$${\left({0},{3}\right)}$$$.

If $$${y}={0}$$$ then $$${3}{x}+{2}\cdot{0}={6}$$$ or $$${x}={2}$$$. Therefore, second point is $$${\left({2},{0}\right)}$$$.

Now draw two found points and line through them.

Graph is shown on figure.

Suppose that we have two lines $$${y}={m}_{{1}}{x}+{b}_{{1}}$$$ and $$${y}={m}_{{2}}{x}+{b}_{{2}}$$$.

Lines are **parallel** if $$${m}_{{1}}={m}_{{2}}$$$. If in addition $$${b}_{{1}}={b}_{{2}}$$$ then lines are same.

Lines are **perpendicular** if $$${m}_{{1}}{m}_{{2}}=-{1}$$$.

**Example 3**. Find equation of line that is perpendicular to line $$${4}{x}+{3}{y}={2}$$$ and passes through point $$${\left(-{1},{3}\right)}$$$.

Let's rewrite given line in standard form: $$${y}=-\frac{{4}}{{3}}{x}+\frac{{2}}{{3}}$$$.

If slope of required line is $$${m}$$$, then since lines are perpendicular we have that $$${m}\cdot{\left(-\frac{{4}}{{3}}\right)}=-{1}$$$ or $$${m}=\frac{{3}}{{4}}$$$.

Thus, equation of line is $$${y}=\frac{{3}}{{4}}{x}+{b}$$$. To find constant $$${b}$$$ we use the fact that line passes through point $$${\left(-{1},{3}\right)}$$$: $$${3}=\frac{{3}}{{4}}\cdot{\left(-{1}\right)}+{b}$$$ or $$${b}=\frac{{15}}{{4}}$$$.

So, equation of line is $$${y}=\frac{{3}}{{4}}{x}+\frac{{15}}{{4}}$$$ or $$${4}{y}-{3}{x}={15}$$$.