# What is the equation of the line with slope  m= 14/25  that passes through  (23/5, (-23)/10) ?

May 7, 2016

$y = \frac{14 x}{25} + 4 \frac{219}{250}$

This is a somewhat unrealistic question, and becomes an exercise in arithmetic rather than maths.

#### Explanation:

There are 2 methods:

Method 1 . uses the formula $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
This is great to use if you know the slope (m) and one point, which is exactly what we have here. It involves one step of substitution and a bit of simplifying.

$\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$

$\left(y - \left(- \frac{23}{10}\right)\right) = \frac{14}{25} \left(x - \frac{23}{5}\right)$

$y + \frac{23}{10} = \frac{14 x}{25} - \frac{14}{25} \times \frac{23}{5} \text{ } \times 250$

$250 y + 250 \times \frac{23}{10} = 250 \times \frac{14 x}{25} - 250 \times \frac{14}{25} \times \frac{23}{5}$

$250 y + 575 = 140 x - 28 \times 23$

$250 y = 140 x + 1219$

$y = \frac{14 x}{25} + 4 \frac{219}{250}$

Method 2 uses $y = m x + c$

Subst for $m , x \mathmr{and} y$ to find $c$

$\left(- \frac{23}{10}\right) = \frac{14}{25} \times \frac{23}{5} + c \text{ } \times 250$

$250 \times \left(- \frac{23}{10}\right) = 250 \times \frac{14}{25} \times \frac{23}{5} + 250 c$

$- 575 = 644 + 250 c$

$1219 = 250 c$

$c = \frac{1219}{250} = 4 \frac{219}{250}$

This leads to the same equation, using values for m and c.

$y = \frac{14 x}{25} + 4 \frac{219}{250}$.