How do you find a standard form equation for the line with $A (- 1, 4)$ $A (-1,4)$ ; Slope: $\frac{2}{5}$ $2/5$ ?

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Answer 1 · 2017-07-19T20:29:41

$y - \frac{2}{5} x = 4 \frac{2}{5}$ $y-2/5x=4frac(2)(5)$ which leads to

$2 x - 5 y = - 22$ $2x-5y =-22$

The standard form of a line is expressed in the following form
$A x + B y = C$ $Ax+By=C$ , where $A, B and C$ $A,B and C$ are integers

To find this form use the formula $(y - y_{1}) = m (x - x_{1})$ $(y-y_1)=m(x-x_1)$

substitute in the point and slope (m)

$(y - 4) = \frac{2}{5} (x - (- 1))$ $(y-4)=2/5(x-(-1))" "$ or $(y - 4) = \frac{2}{5} (x + 1)$ $" "(y-4)=2/5(x+1)$

$y - 4 = \frac{2}{5} x + \frac{2}{5}$ $y-4=2/5x+2/5$

$y = \frac{2}{5} x + \frac{2}{5} + 4$ $y=2/5x+2/5+4$

$y = \frac{2}{5} x + 4 \frac{2}{5}$ $y=2/5x+4frac(2)(5)" "$ now put into standard form

Multiply by $5$ $5$ to clear the denominators

$5y-(cancel5xx2)/cancel5x=cancel5xx22/cancel5$

$5y-2x=22$

$rArr 2x-5y = -22$

How do you find a standard form equation for the line with A (-1,4)A(−1,4)A (-1,4); Slope: 2/5252/5?