Lines...

**classicstrings** · March 25th 2006, 04:22 AM

Hi all! My friend recommended this site for me! Hope you can help me out!

I have a few quesions:

1.Find the angle of the line joining the points (-1,2) and (4,6), makes with the x axis.

2.Find the equation of the line passing through (-1,2) and making an angle of 53 degrees and 8 minutes with the x-axis.

3.Find the equation of the line containing the point (2,-5) that is
a) parallel to the line y=-2x+7
b) Perpendicular to the line y = -2x +7

4. Find the values of a and b for which x^3 + ax^2 - 10x + b is divisible by x^2 + x - 12.

THanks a lot for your help!

**earboth** · March 25th 2006, 05:04 AM

Originally Posted by classicstrings

Hi all! My friend recommended this site for me! Hope you can help me out!

I have a few quesions:

1.Find the angle of the line joining the points (-1,2) and (4,6), makes with the x axis.

2.Find the equation of the line passing through (-1,2) and making an angle of 53 degrees and 8 minutes with the x-axis.

3.Find the equation of the line containing the point (2,-5) that is
a) parallel to the line y=-2x+7
b) Perpendicular to the line y = -2x +7

4. Find the values of a and b for which x^3 + ax^2 - 10x + b is divisible by x^2 + x - 12.

THanks a lot for your help!

Hello,

to 1.: Use the slope-formula:
$m=\frac{y_1-y_2}{x_1-x_2}$ . Plug in the values you know and you'll get:
$m=\frac{2-6}{(-1)-4}=\frac{-4}{-5}$
The slope is the tangens of the angle between line and x-Axis. So the angle is the arctan(m)
$arctan\left(\frac{4}{5}\right)\approx 38^\circ39'35.31''$

to 2.: Even if it isn't exact I'll use $m=tan(53^\circ8')=\frac{4}{3}$
Now use the point-slope-formula:
$m=\frac{y-y_1}{x-x_1}$ . Plug in the values you know:
$\frac{4}{3}=\frac{y-2}{x-(-1)}$ . Solve for y and you'll get:
$y=\frac{4}{3} \cdot (x+1)+2=\frac{4}{3} \cdot x+\frac{10}{3}$

to 3.: a) The slope is m = -2. Now use the point-slope-formula and you'll get:y=-2x-1
b) Two lines are perpendicular if $l_1 \bot l_2 \Longleftrightarrow m_1 \cdot m_2=-1$ where l means a line and m the slope of the corresponding line.
Therefore the perpendicular slope is $m_p=-\frac{1}{-2}=\frac{1}{2}$ . Now use the point-slope-formula and you'll get: $y=\frac{1}{2} \cdot x - 6$

to 4.: Do the long division $\left(x^3+ax^2-10x+b \right) / \left(x^2+x-12 \right)$
The result is $x+a-1+\frac{(3-a)x+b+12a-12}{x^2+x-12}$
Divisible means the remainder must be zero(0):
$3-a=0\ \wedge \ b+12a-12=0$ . From the 1rst equation you'll get a=3. Plug this value into the 2nd equation and you'll get b = -24.

Greetings

EB

**classicstrings** · March 25th 2006, 06:46 PM

Thanks earboth! However, I cant seem to get my head around the last question and its long division. I can do normal LD with numbers but i get confused with letters. Can you help me please?

**ThePerfectHacker** · March 25th 2006, 07:05 PM

Originally Posted by classicstrings

Thanks earboth! However, I cant seem to get my head around the last question and its long division. I can do normal LD with numbers but i get confused with letters. Can you help me please?

What earboth is trying to say, is that whenver you preform long division and you CAN divide one polynomial by another then the remainder after long division is zero.

Earboth, preformed long division and he got (remainder),
$(3-a)x+b+12a-12$
But, since you said you want to be able to divide those two polynomials evenly your remainder, needs to be zero.
Notice that this remainder is itself a polynomial and you want it to be zero. Whenever, you have a polynomial equivalent to zero its corresponding parts (coefficients) must be zero. Since the parts in this remainder are,
$(3-a)x+(b+12a-12)$ , since you want this to be zero its x coefficient is,
$3-a=0$
and its constant coefficient is,
$b+12a-12=0$
If you solve this you get,
$(a,b)=(3,-24)$

**classicstrings** · March 25th 2006, 07:09 PM

PerfectHacker, i understand that part, what i dont understand is long division with letters instead of actual number values. I dont know how to do that. Thx

Sorry for the trouble im causing you!

**classicstrings** · March 26th 2006, 04:54 AM

I was hoping someone might be able to show me how this:

$ \left(x^3+ax^2-10x+b \right) / \left(x^2+x-12 \right) $

is done?

Thanks a bunch!

**TD!** · March 26th 2006, 05:19 AM

Originally Posted by classicstrings

I was hoping someone might be able to show me how this is done:

$ \left(x^3+ax^2-10x+b \right) / \left(x^2+x-12 \right) $

You could do the long polynomial division and then express that the rest has to be zero, this will give conditions on a and b. A rather annoying task if you ask me...

You can also see that the denominator can be factorized into (x-3)(x+4). So for the divisibility, you'd want the numerator to have these factors as well. In other words, you can x = 3 and x = -4 to be zeroes of the numerator. Plugging these values in the numerator and letting the results equal 0 give two simple eqaution in a and b.

These form a lineair system which can be easily solved, you should find a = 3 and b = -24, check that yourself

**earboth** · March 26th 2006, 10:10 AM

Originally Posted by classicstrings

I was hoping someone might be able to show me how this:
$ \left(x^3+ax^2-10x+b \right) / \left(x^2+x-12 \right) $
is done?

Thanks a bunch!

Hello,

you said that you know how to do the longdivision with numbers. So I'll try to explain the long division with terms. There is only one problem: I translate my explanations (which are in German of course) into English, but I'm not certain if it is understandable English what I produce. So it could happen that you don't understand my explanation because my English is in a desastrous state.

Division means that you have to subtract manifolds of the divisor from the dividend. You have to calculate those manifolds you have to subtract:

$\underbrace{(x^3+ax^2-10x+b)}_\csub{dividend} / \underbrace{(x^2+x-12)}_\csub{divisor}$

1rst step: Divide the first summand of the dividend by the first summand of the divisor and you get x. That's the first result. Now multiply the divisor by x and you get
$x^3+x^2-12x$ . Substract this manifold of the divisor from the dividend:

$\ (x^3+ax^2-10x+b)} / (x^2+x-12)}=x$
$-(x^3+x^2-12x)$ you'll get:
$\ (ax^2-x^2-10x+12x+b)} / (x^2+x-12)}=x$
$\ ((a-1)x^2+2x+b)} / (x^2+x-12)}=x$

2nd step: Divide the first summand of the dividend by the first summand of the divisor. You get (a-1). That's the 2nd result. Now multiply the divisor by (a-1) and substract this manifold of the divisor from the divisor:
$\ ((a-1)x^2+2x+b)} / (x^2+x-12)}=x+(a-1)$
$-((a-1)x^2+(a-1)x-(a-1) \cdot 12)$
$2x-(a-1)x+b+(a-1) \cdot 12$
$(3-a)x+b+12a-12$ That's the remainder of the division because there isn't any $x^2$ left in the dividend to divide by the $x^2$ of the divisor.

You already know what has happened to the remainder.

I would appreciate a short notice if you could understand this text. Only to be sure that I don't write too much nonsense.

Greetings

EB

**classicstrings** · March 26th 2006, 08:41 PM

Thanks a lot earboth for taking the time to type all that for me! Yes i understand it after a while looking at it and working it.

Thread: Lines...

LinkBack

Thread Tools

Display

Lines...

Similar Math Help Forum Discussions

Lines of best fit

Distance between 2 parametric lines - Vectors/Lines and Planes

Geometry problem prove lines are parallel, cyclic quadrilaterals,other parallel lines

Tangent Lines, Parallel Lines, and Normal Lines

Lines

Search Tags