What is the equation of the line with slope $m = - \frac{31}{25}$ $m= -31/25$ that passes through $(- \frac{6}{5} \frac{11}{10})$ $(-6/5 11/10)$ ?

Question

1489 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T13:02:39

$155 x + 125 y - 939 = 0$ $color(purple)(155x + 125y -939 = 0$

Given : $m = - (\frac{31}{25}), x_{1} = - (\frac{6}{5}), y_{1} = (\frac{11}{10})$ $m = -(31/25), x_1 = -(6/5), y_1 = (11/10)$

Slope point form of equation is $y - y_{1} = m (x - x_{1})$ $y - y_1 = m (x - x_1)$

$y - (\frac{11}{10}) = - (\frac{31}{25}) \cdot (x + (\frac{6}{5})$ $y - (11/10) = - (31/25) * (x + (6/5)$

$25 y - 225 = - 31 x - \frac{186}{5}$ $25y - 225 = -31x - 186/5$

$125 y - 1125 = - 155 x - 186$ $125y - 1125 = -155x - 186$

$155 x + 125 y - 939 = 0$ $color(purple)(155x + 125y -939 = 0$

graph{(-155x + 939) / 125 [-10, 10, -5, 5]}

What is the equation of the line with slope m=−3125 m= -31/25 that passes through (−651110) (-6/5 11/10) ?