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Write the equation of the line which passes through $(1, –2)$ and is perpendicular to the line with equation $5y – x = 1$.

I know that I need to put the equation into slope-intercept form but what is the step that I take after this.

$$y = \frac 1 5 x + \frac 1 5$$

I read something about the negative reciprocal but that isn't working for me.


Also, here are the possible answers:

a. $5x + y = 3$ ANSWER
b. $5x – y = 1$
c. $x + 5y = –2$
d. $x – 5y = 3$

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We know that if two perpendicular lines have gradients $m_1$ and $m_2$ then $m_1 m_2 = -1$. As you've put the line you're given in slope-intercept form you know that it has gradient $\frac{1}{5}$. This tells you that the line you're trying to find has gradient $m$ satisfying $\frac{m}{5} = -1$ which gives $m = -5$.
So we know now that the line you're trying to find is of the form $y = -5x + c$ for suitable $c$ in slope-intercept form. But we know it passes through (1,-2) so we substitute in to get $-2 = -5 + c$ giving $c = 3$. So the line is $y = -5x + 3$, this rearranges to answer a.

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If two lines with slopes $m$ and $m'$ respectively are perpendicular to each other we have:

$$ mm'=-1 $$

In our case, we're looking for $m'$:

$$ m'/5 = -1\\ m' = -5 $$ So our line equation becomes:

$$ y = -5x+p $$

Now using that the lines passes from $(1,-2)$, we solve for $p$:

$$ -2 = -5+p\\ p = 3 $$

Our equation becomes:

$$ y = -5x+3\\ y+5x-3 = 0 $$

So the answer is $a$

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Hint:

A line perpendicular to $ax+by=c$ takes the form $bx-ay=d$.

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Notice the slope of line perpendicular to the line: $5y-x=1\iff y=\frac{1}{5}x+\frac{1}{5}$ $$=\frac{-1}{\frac{1}{5}}=-5$$

Hence, the equation of the line passing through the point $(1, -2)$ & having slope $\frac{1}{5}$ $$y-(-2)=-5(x-1)$$ $$\color{blue}{5x+y=3}$$

Hence, option (a) is correct.

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The vector $(-1,5)$ is normal to the given line. Hence $(5,1)$ is normal to the perpendicular. If this perpendicular passes through the point $(1,-2)$, its equation is $$5x+y=5\cdot \color{red}1 +(\color{red}{-2})=3.$$

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