# What is the equation of the line that passes through (5, -2) and (3, 4)?

Aug 13, 2018

$y = - 3 x + 13$

#### Explanation:

$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.

•color(white)(x)y=mx+c

$\text{where m is the slope and c the y-intercept}$

$\text{to calculate m use the "color(blue)"gradient formula}$

•color(white)(x)m=(y_2-y_1)/(x_2-x_1)

$\text{let "(x_1,y_1)=(5,-2)" and } \left({x}_{2} , {y}_{2}\right) = \left(3 , 4\right)$

$m = \frac{4 - \left(- 2\right)}{3 - 5} = \frac{6}{- 2} = - 3$

$y = - x + c \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$

$\text{to find c substitute either of the 2 given points into}$
$\text{the partial equation}$

$\text{using "(3,4)" then}$

$4 = - 9 + c \Rightarrow c = 4 + 9 = 13$

$y = - 3 x + 13 \leftarrow \textcolor{red}{\text{is the equation of the line}}$

Aug 13, 2018

The eqn. of $\text{line passes through } \left(5 , - 2\right) \mathmr{and} \left(3 , 4\right)$ is

$3 x + y - 13 = 0$

#### Explanation:

The eqn. of line passes through $A \left({x}_{1} , {y}_{1}\right) \mathmr{and} B \left({x}_{2} , {y}_{2}\right)$ is

$| \left(x , y , 1\right) , \left({x}_{1} , {y}_{1} , 1\right) , \left({x}_{2} , {y}_{2} , 1\right) | = 0$

We have , two points :$A \left(5 , - 2\right) \mathmr{and} B \left(3 , 4\right)$

So, the eqn. of $\text{line passes through } A \mathmr{and} B$ is

$| \left(x , y , 1\right) , \left(5 , - 2 , 1\right) , \left(3 , 4 , 1\right) | = 0$

Expanding we get

$x \left(- 2 - 4\right) - y \left(5 - 3\right) + 1 \left(20 + 6\right) = 0$

$\therefore x \left(- 6\right) - y \left(2\right) + 1 \left(26\right) = 0$

$\therefore - 6 x - 2 y + 26 = 0$

Divding each term by $\left(- 2\right)$

$\therefore 3 x + y - 13 = 0$

$\therefore$The eqn. of $\text{line passes through } \left(5 , - 2\right) \mathmr{and} \left(3 , 4\right)$ is

$3 x + y - 13 = 0$

Aug 14, 2018

$y = - 3 x + 13$

#### Explanation:

There is a useful formula which can be used to find the equation of a line if two points are known. It is based on the slope formula.

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

The points are $\left(5 , - 2\right) \mathmr{and} \left(3 , 4\right)$

$\frac{y - \left(- 2\right)}{x - 5} = \frac{4 - \left(- 2\right)}{3 - 5}$

$\frac{y + 2}{x - 5} = \frac{4 + 2}{3 - 5} = \textcolor{b l u e}{\frac{6}{-} 2 = - \frac{3}{1}} \text{ } \leftarrow$ (this is the slope)

$\frac{y + 2}{x - 5} = - \frac{3}{1}$

$y + 2 = - 3 \left(x - 5\right)$

$y = - 3 x + 15 - 2$

$y = - 3 x + 13$