Normal lines.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Hello.

I have a problem solving a task, which I've done some others similar not long time ago. But I need some help here...

$$\text{Is given:}$

$$\text{What should be the value of "m" so the line: }\;\;(m+3)x-(m-2)y-5=0\;\;\;$

$$\;\;\;\text{would be normal with line:}\;\;y=x+3\;\;$

Attempted:

$(m+3)x-(m-2)y=5

$$y=\frac{-(m+3)x}{-(m-2)}+\frac{5}{-(m-2)}$

$$y=\frac{(m+3)}{(m-2)}x+\frac{5}{(m-2)}$

Now here's the problem, if I want to take k which in this case should be: $$k=\frac{m+3}{m-2}$ the denominator of fraction $$\frac{5}{m-2}$ contains letter m
Which one we need to calculate the value of m, anyway, my other similar tasks I had before, always was like... a line at one point... example:

Point $A(1,4)

normal with line x+2y-2=0

So I just used:

$$y-y_0=k(x-x_0)\;\;\;\text{and}\;\;\;k\cdot k_1=-1$

But in this case I don't have any point, I don't know what should I do, any help would be great, Thank you. (Sorry for my bad english

)

The multiple choices are:

$$A)\;m=-\frac12$

$$B)\;m=-\frac32$

$$C)\;m=\frac12$

$$D)\;m=\frac32$

royhaas · **Posted:** Mon, 27 Feb 2012 18:56:29 UTC

The slope needs to be -1.

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

royhaas wrote:

The slope needs to be -1.

How do you mean -1 ?

kreshnik · **Joined:** Sun, 25 Dec 2011 00:28:25 UTC **Posts:** 95

Evaluating the problem I think I just found something...

$$\text{Let's take}\;\; k=\frac{m+3}{m-2}\;\;\;\text{and we know that in this case we have:}\;\;k \cdot k_1=-1\;\;\text{Hence:}$

From lines:

$$y=\left(\frac{m+3}{m-2}\right)x+\frac{5}{m-2}\;\;\text{and}\;\;y=x+3\;\;\text{we have:}$

$$k=\frac{m+3}{m-2}\;\;\;\text{And}\;\;\;k_1=1$

Therefore:

$$\left(\frac{m+3}{m-2}\right)\cdot 1=-1$

$$(m-2)\cdot (-1)=m+3$

$$2-m=m+3\;\;\;\Longrightarrow\;\;\;2m=-1\;\;\;\Longrightarrow\;\;\boxed{m=-\frac12}$

Let's be sure...

We have:

$(m+3)x-(m-2)y-5=0

Substituting we get:

$$\left(-\frac12+3\right)x-\left(-\frac12-2\right)y-5=0$

$$\left(\frac52\right)x+\left(\frac52\right)y=5$

$$\left(\frac52\right)y=5-\frac52x$

$$y=\frac{-\frac{5}{2}}{\frac52}x+\frac{5}{\frac52}$

$y=-x+2

Getting the lines:

$$y=-x+2\;\;\;\text{and}\;\;\;y=x+3$

We have:

$$-1\cdot 1=-1$

True...or not??

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Normal lines.

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