Author Topic: Samuch's facepalm question thread (Read 2282 times) Tweet Share

samuch · « **on:** December 28, 2009, 09:08:39 pm »

i guess its because im in holiday mode (i really hope thats the reason :buck2:) but i dont get how to make x the subject of this:

$\frac{p-qx}{t}+p=\frac{qx-t}{p}$

i end up with t/q which is really wrong...

also

Three points have coordinates A(1,7), B(7,5) and C(0,-2). Find:

a) the equation of the perpendicular bisector of AB
b) the point of intersection of this perpendicular bisector and BC

iv done some of part a... iv gotten 3 as the gradient of the perpendicular line but cant get the y-int, forgotten how to

TrueTears · « **Reply #1 on:** December 28, 2009, 09:21:48 pm »

$\frac{p-qx}{t}+p=\frac{qx-t}{p}$

$\frac{p-qx}{t} + \frac{pt}{t} = \frac{qx-t}{p}$

$(p-qx+pt)p=(qx-t)(t)$

$p^2-pqx+p^2t = qxt-t^2$

$p^2+p^2t+t^2=qxt+pqx$

$\frac{p^2+p^2t+t^2}{qt+pq}=x$

Quote

Three points have coordinates A(1,7), B(7,5) and C(0,-2). Find:

a) the equation of the perpendicular bisector of AB
b) the point of intersection of this perpendicular bisector and BC

a) First find the equation of the line AB.

Then you need to find the midpoint of AB. (That is what "bisector" means)

Then you need to find the equation of the line that passes through this midpoint also this line must be perpendicular to the line AB, ie $\text{gradient}_{\text{perpendicular line}} \times \text{gradient}_{\text{AB}} = -1$

b) First you need to find the equation of the line BC and then you need to set up a simultaneous equation to solve for the intersection point.

zzdfa · « **Reply #2 on:** December 28, 2009, 09:22:40 pm »

$p^2-qpx+p^2t=qxt-t^2$ multiply both sides by pt

$qxt+qpx=p^2-p^2t+t^2$ rearrange

$x(qt+qp)=p^2+p^2t+t^2$ factorize

samuch · « **Reply #3 on:** December 28, 2009, 09:25:20 pm »

oh

thanks both of you
hmm TT i sent you a PM earlier today, not sure if you checked

TrueTears · « **Reply #4 on:** December 28, 2009, 09:26:43 pm »

Quote from: samuch on December 28, 2009, 09:25:20 pm

oh thanks both of you
hmm TT i sent you a PM earlier today, not sure if you checked

Lol sorry, someone else just sent me a pm and it went to the next page of pm's, so didn't see yours which was on the other page, sorry!

Lighties · « **Reply #5 on:** December 28, 2009, 09:27:42 pm »

Just wondering, how do you do the spiffy thing where you set out the equations neatly (the format)?

TrueTears · « **Reply #6 on:** December 28, 2009, 09:28:27 pm »

Quote from: Lighties on December 28, 2009, 09:27:42 pm

Just wondering, how do you do the spiffy thing where you set out the equations neatly (the format)?

Do you mean $\LaTeX$ ?

Have a look here http://vcenotes.com/forum/index.php/topic,10280.0.html

Lighties · « **Reply #7 on:** December 28, 2009, 09:31:37 pm »

Yeah, I did. Thank you! (:

It looks more complicated than I thought.

TrueTears · « **Reply #8 on:** December 28, 2009, 09:32:49 pm »

Quote from: Lighties on December 28, 2009, 09:31:37 pm

Yeah, I did. Thank you! (:

It looks more complicated than I thought.

It just takes some time to learn the code etc, you will get used to it soon if you practise

TrueTears · « **Reply #9 on:** December 28, 2009, 09:37:10 pm »

Quote from: samuch on December 28, 2009, 09:08:39 pm

iv done some of part a... iv gotten 3 as the gradient of the perpendicular line but cant get the y-int, forgotten how to

Yeah that's to find the value of $c$ in $y = mx+c$

But... $y -y_1 = m(x-x_1)$ comes to rescue! ( $(x_1,y_1)$ is the point where the line passes through, so just sub it in

)

samuch · « **Reply #10 on:** December 28, 2009, 09:40:20 pm »

Quote from: TrueTears on December 28, 2009, 09:37:10 pm

Quote from: samuch on December 28, 2009, 09:08:39 pm
iv done some of part a... iv gotten 3 as the gradient of the perpendicular line but cant get the y-int, forgotten how to
Yeah that's to find the value of $c$ in $y = mx+c$

But... $y -y_1 = m(x-x_1)$ comes to rescue! ( $(x_1,y_1)$ is the point where the line passes through, so just sub it in )

okok so maybe that equation is better

samuch · « **Reply #11 on:** December 30, 2009, 03:19:48 pm »

another one

simultaneous equations, solve for x and y:

$(a+b)x+cy=bc$

$(b+c)y+ax=-ab$

Stroodle · « **Reply #12 on:** December 30, 2009, 04:52:03 pm »

Quote from: samuch on December 30, 2009, 03:19:48 pm

another one

simultaneous equations, solve for x and y:

$(a+b)x+cy=bc$

$(b+c)y+ax=-ab$

From the first equation you can see that $x=\frac{bc-cy}{a+b}$

Plugging this value of $x$ into the second equation yields:

$(b+c)y+\frac{abc-acy}{a+b}=-ab$

Then after making the common denominator $a+b$ and expanding you get:

$-aby-b^2y-bcy-abc=a^2b+ab^2$

$-y(ab+b^2+bc)=a(ab+b^2+bc)$

$\therefore y=-a$

You can now plug this value of $y$ into either of the first two equations to solve for $x$

Edit* Another way that may be easier would be to multiply the first equation by $a$ and the second by $(a+b)$ , then subtract the second equation from the first etc..

samuch · « **Reply #13 on:** December 30, 2009, 04:55:56 pm »

^ ah i get it

thank you!
looked harder than it was

samuch · « **Reply #14 on:** January 15, 2010, 01:39:45 pm »

another noob question:

find the coordinates of the points of intersection

x^2 + y^2 = 178
x + y = 13

and

x + y = 28
xy = 187

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