# How do you write an equation in standard form when given slope and point on line (4, 1), m = -1/2?

Sep 14, 2016

The equation of the line is $y = - \frac{1}{2} x + 3$

#### Explanation:

Let the equation of line is y=mx+c or y= -1/2x+c ; (4,1) is on the line. So $1 = - \frac{1}{2} \cdot 4 + c \mathmr{and} c = 3$. Hence the equation of the line is $y = - \frac{1}{2} x + 3$ [Ans]

Sep 14, 2016

$y = - \frac{1}{2} x + 3$

#### Explanation:

We can write the equation in $\textcolor{b l u e}{\text{point-slope form}}$

That is $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$

here $m = - \frac{1}{2} \text{ and } \left({x}_{1} , {y}_{1}\right) = \left(4 , 1\right)$

substitute these values into the equation.

$y - 1 = - \frac{1}{2} \left(x - 4\right)$

distribute the bracket and simplify.

$y - 1 = - \frac{1}{2} x + 2$

$y = - \frac{1}{2} x + 2 + 1$

$\Rightarrow y = - \frac{1}{2} x + 3 \text{ is equation in standard form}$