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# Linear equations [Formulas and examples]

In this article we use the most common form of linear equation where m is slope and c is y-intercept.

$$y=mx+c$$

If two points $A(x_1,y_1)$ and $B(x_2,y_2)$ are belonging to linear equation we can easily find slope with this formula below.

$$m=\frac{y_2-y_1}{x_2-x_1}$$

If two linear equations are parallel to each other then their slopes are the same $m_1=m_2$. And if two linear equations are perpendicular to each other then their slopes satisfy this rule $m_1\cdot m_2=-1$ requirements.

## Solved examples

Write linear function equation, if $m = 0.5$ and point with coordinates $(3,12)$ belongs to this function. How solve this task?

General equation of the straight line is $y=mx+c$. If $m = 0.5$, then we have $y=0.5x+c$. Now we can easy find c by inserting given point coordinates to our equation.

$$12=0.5 \cdot 3 + c, \ \ c=10.5$$

Answer: $y=0.5x+10.5$

Find the slope of the function, when points $A(1,3)$ and $B(7,12)$ belongs to this function?

If we have two points of the straight line function, we can use this formula $\displaystyle\frac{y_2-y_1}{x_2-x_1}$ to find the slope of this line. So there we have:

$$\text{Slope}=\frac{y_2-y_1}{x_2-x_1}=\frac{12-3}{7-1}=\frac{9}{8}=1\frac{1}{8}$$

Find the equation of the line that passes though the points $(-2,5)$ and $(8,15)$ ?

General equation of the straight line function is $y=mx+c$, where m is slope.

$$m=\frac{y_2-y_1}{x_2-x_1}=\frac{15-5}{8-(-2)}=\frac{10}{10}=1$$

If $m = 1$, then we have $y=x+c$. Now we can easy find c by inserting one of the given points coordinates to our equation.

$$15= 8 + c, \ \ c=7$$

Answer: $y=x+7$

Find the equation of the line that is parallel to $y = 4x - 1$ and passes though the point $(7,30)$ ?

Parallel straight lines have the same slopes $m_1=m_2$. General equation of the straight line is $y=mx+c$. If $m = 4$, then we have $y=4x+c$. Now we can easy find c by inserting given point coordinates to our equation.

$$30=4 \cdot 7 + c, \ \ c=2$$

Answer: $y=4x+2$

Find the equation of the line that is perpendicular to the $y = 4x - 1$ and passes through the point $(8,5)$ ?

General equation of the straight line is $y=mx+c$. Characteristic of perpendicular straight lines slopes: $m_1\cdot m_2=-1$. Now we can find the lope of our equation:

$$m\cdot 4 = -1, \ \ m=-\frac{1}{4}$$

If $m=-\frac{1}{4}$, then we have $y=-\frac{1}{4}x+c$. Now we can easy find c by inserting given point coordinates to our equation.

$$5=-\frac{1}{4} \cdot 8 + c, \ \ c=7$$

Answer: $y=-\frac{1}{4}x+7$