What is the equation of the line between $(1, 3)$ $(1,3)$ and $(- 4, 1)$ $(-4,1)$ ?

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Answer 1 · 2016-06-30T09:09:24

Its equation is ; $5 y = 2 x + 13$ $5y = 2x + 13$

first you will find its gradient ( slope );

gradient = $\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$ $(y_1 - y_2)/(x_1-x_2)$ or $\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ $(y_2 - y_1)/(x_2-x_1)$

{I am applying the first one]

gradient = $\frac{3 - 1}{1 - (- 4)}$ $( 3-1)/(1-(-4))$

gradient = $\frac{2}{5}$ $2/5$

now put gradient , in this equation

y = m*x + c

{where m is gradient & c is y-intercept (where x is 0) }

$y = (\frac{2}{5}) \cdot x + c$ $y = (2/5)*x + c$

{now put the value of y and x, either use first point or the second one i.e. (1, 3) or (-4, 1) }

$3 = (\frac{2}{5}) \cdot 1 + c$ $3 = (2/5)* 1 +c$
$3 - (\frac{2}{5}) = c$ $3-(2/5) = c$
$\frac{13}{5} = c$ $13/5 = c$

{c is found now put m and c in the equation y = mx + c}

$y = (\frac{2}{5}) \cdot x + \frac{13}{5}$ $y = (2/5)*x + 13/5$

$5 y = 2 x + 13$ $5 y= 2x + 13$

{you can put any two points and the answer will be the same}

equation found...!

What is the equation of the line between (1,3)(1,3) and (−4,1)(-4,1)?