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The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is

  1. Perpendicular to line $A$.

  2. Parallel to line $A$.

Attempting to find the parallel,

I tried $$y = -\frac{1}{2}x + \frac{1}{6}$$

$$y - (1) = -\frac{1}{2}(x-5)$$

$$Y = -\frac{1}{2}x - \frac{1}{10} - \frac{1}{10}$$

$$y = -\frac{1}{2}x$$

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  • $\begingroup$ Do you know the relation between slope of two lines when they are parallel and perpendiculer to each other? $\endgroup$ – Chiranjeev_Kumar Aug 10 '15 at 17:42
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Notice, in general, the equation of the line passing through the point $(x_1, y_1)$ & having slope $m$ is given by the point-slope form: $$\color{blue}{y-y_1=m(x-x_!)}$$

We have, equation of line A: $3x+6y-1=0\iff \color{blue}{y=-\frac{1}{2}+\frac{1}{6}}$ having slope $-\frac{1}{2}$

1.) slope of the line passing through $(5, 1)$ & perpendicular to the line A $$=\frac{-1}{\text{slope of line A}}=\frac{-1}{-\frac{1}{2}}=2$$ Hence, the equation of the line: $$y-1=2(x-5)$$ $$y-1=2x-10$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{2x-y-9=0}}$$

2.) slope of the line passing through $(5, 1)$ & parallel to the line A $$=\text{slope of line A}=-\frac{1}{2}$$ Hence, the equation of the line: $$y-1=-\frac{1}{2}(x-5)$$ $$2y-2=-x+5$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{x+2y-7=0}}$$

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For the second case:

the line you want to find its equation is parallel to $A$, so it means they have the same slope$-\frac{1}{2}$, which yields:

$$ {y}_{1} = -\frac{1}{2}x+p $$

Now you'll find $p$ from another condition you gave(the line passes from $(5,1)$), so we get :

$$ 1 = \frac{-5}{2}+p\\ p = 1+\frac{5}{2} = \frac{7}{2} $$

We finally get:

$$ {y}_{1} = -\frac{1}{2}x+\frac{7}{2} $$

For the first question you should know that if $m$ is the slope of line $A$ and $m'$ is the slopeof line $B$, and $A$ and $B$ are perpendicular then we have:

$$ mm' = -1 $$

For our case we have $m = 1/2$, so $m' = 2$ and we get:

$$ {y}_{2} = 2x+p' $$

Can you continue from here?

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    $\begingroup$ So the perpendicular would be y=2x-9? $\endgroup$ – Madison Aug 10 '15 at 20:10
  • $\begingroup$ Yes you're correct ^^ $\endgroup$ – Oussama Boussif Aug 10 '15 at 20:36
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Line perpendicular to A that passes through the point $(5,1)$: it is directed by the normal vector to A: $(3,6)$ \ or $(1,2)$. Hence its equation is $$\frac{x-5}1=\frac{y-1}2\iff 2x-y-9=0. $$

Line parallel to A:it has the smame normal vector as A: $$3x+6y=3\cdot 5+6\cdot 1\iff 3x+6y-21=0.$$

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