# Find the equations of line joining points (-2,3) and (1,4)?

Nov 15, 2016

The equation for line joining two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by $\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$ and for given points it is $y = \frac{1}{3} x + \frac{11}{3}$

#### Explanation:

Let the slope intercept form of equation be $y = m x + c$

here we do not know the slope $m$ and $y$-intercept $c$

What we know is that this passes through the two coordinate pairs, say $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$.

As such we have three equations

$y = m x + c$ ......(1)

${y}_{1} = m {x}_{1} + c$ ......(2) and

${y}_{2} = m {x}_{2} + c$ ......(3)

Now using these let us eliminate $m$ and $c$

subtracting (2) from (1), we get $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$ ......(4)

and subtracting (2) from (3), we get $\left(y - 2 - {y}_{1}\right) = m \left({x}_{2} - {x}_{1}\right)$ ......(5)

Dividing (4) by (5)

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

As the two points are $\left(- 2 , 3\right)$ and $\left(1 , 4\right)$, the equation is

$\frac{y - 3}{4 - 3} = \frac{x - \left(- 2\right)}{1 - \left(- 2\right)}$

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

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

or $3 y - 9 = x + 2$

or $3 y = x + 11$ or $y = \frac{1}{3} x + \frac{11}{3}$

Nov 15, 2016

For any point on the line, the coordinate pair in slope-intercept form is$\left(x , \frac{x}{3} + \frac{11}{3}\right)$

#### Explanation:

For slope m and intercept c, the equation is y = m x +c.

The slope intercept form for coordinates is (x, m x +c ).

The slope of the line through the given points is

$m = \frac{4 - 3}{1 - \left(- 2\right)} = \frac{1}{3}$

Also, from (1, 4). 4 = i/3(1) + c. So, c = 11/3.

So, the answer is $\left(x , \frac{1}{3} x + \frac{11}{3}\right)$.

Nov 15, 2016

The equation of the line is:

$y = \frac{1}{3} x + \frac{11}{3}$ which can be written as $y = \frac{1}{3} x + 3 \frac{2}{3}$

#### Explanation:

If you are given the coordinates of 2 points on a line, substituting them into the formula below allows you to find the equation immediately. In the process you also calculate the slope.

$\left(- 2 , 3\right) \mathmr{and} \left(1 , 4\right)$
$\left({x}_{1} , {y}_{1}\right) \mathmr{and} \left({x}_{2} , {y}_{2}\right)$

$\textcolor{red}{\frac{y - {y}_{1}}{x - {x}_{1}} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}} \text{ } \leftarrow R H S = m$

$\frac{y - 3}{x - \left(- 2\right)} = \frac{4 - 3}{1 - \left(- 2\right)} = \frac{1}{3}$

$\frac{y - 3}{x + 2} = \frac{1}{3} \text{ } \leftarrow$ now cross-multiply

$3 \left(y - 3\right) = x + 2 \textcolor{w h i t e}{\times \times \times \times \times} \mathmr{and} y - 3 = \frac{1}{3} \left(x + 2\right)$

$3 y - 9 = x + 2 \textcolor{w h i t e}{\times \times \times \times \times \times \times \times x} y = \frac{1}{3} x + \frac{2}{3} + 3$

$3 y = x + 11 \textcolor{w h i t e}{\times \times \times \times \times \times \times \times \times x} y = \frac{1}{3} x + 3 \frac{2}{3}$

$y = \frac{1}{3} x + \frac{11}{3}$