What is the an equation of the line that goes through (−1, −3) and is perpendicular to the line $2 x + 7 y + 5 = 0$ $2x + 7y + 5 = 0$ ?

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Answer 1 · 2015-11-27T17:58:19

$2 y = 7 x + 1$ $2y = 7x + 1$

$r : y = a x + b$ $r: y = ax + b$ is perpendicular to $y = \frac{- 5 - 2 x}{7}$ $y = (-5 - 2x) / 7$

$- \frac{1}{a} = - \frac{2}{7}$ $-1/a = -2/7$

$a = \frac{7}{2}$ $a = 7/2$

$(- 1, - 3) \in r \Rightarrow - 3 = \frac{7}{2} \cdot (- 1) + b$ $(-1, -3) in r Rightarrow -3 = 7/2 * (-1) + b$

$b = - 3 + \frac{7}{2} = \frac{1}{2}$ $b = -3 + 7/2 = 1/2$

$r : y = \frac{7}{2} x + \frac{1}{2}$ $r : y = 7/2 x + 1/2$

What is the an equation of the line that goes through (−1, −3) and is perpendicular to the line 2x+7y+5=02x + 7y + 5 = 0?