maths help!!

[mousa]

hey everyone , well i got stuck on those questions , astar, nid.. anybody..
help me out!..plz i need workin out with reasoning..

1, let p,with coorinates (p,q), be a fixed point on the 'curve ' with equation y=mx+c and let Q with coordinates (r,s), be any other point on y=mx+c . use the fact  thar the coordinates of p and Q satisfy the equation y=mx+c to  show that the gradient of PQ is m for all positions of Q.

2 there are some values of  a, b and c for which the equation ax+by+c=0 does not represent a straight line. give an example of such values


3 two lines have equations  y=m1x+c1 and y=m2x+c2 and m1m2=-1. prove that the lines are perpendicular.


Thanks in advance..

[nid404]

Will answer his tom definitely....have an exam tom..so sorry...tom for sure

[astarmathsandphysics]

I answered these questions somewhere else. m1m2=-1 is the def of perpendicular.

[Ghost Of Highbury]

the answer for the third one is dere in the question itself..

if m1*m2 = -1

the lines are perpendicular...

[slvri]

the answer for the third one is dere in the question itself..

if m1*m2 = -1

the lines are perpendicular...

nope the answer aint in the question...........we did this q in schl a few days ago.....the teacher gave us a weird sort of proof....ill post it when i find it

[Ghost Of Highbury]

this is what i remember from my add. math book

"two non-vertical lines with gradients m1 and m2 are perpendicular if m1*m2 = -1"

[nid404]

oh no adi adi it's not as simple as that.....
I'll do it in sometime...I'm just  back from an exam which i am sure to flunk......I mixed my reagents and my practicals was such a mess

[mousa]

oh no adi adi it's not as simple as that.....
I'll do it in sometime...I'm just  back from an exam which i am sure to flunk......I mixed my reagents and my practicals was such a mess

ooohh1!! nid,  chem practical?

[Ghost Of Highbury]

ohh..okk...sure..

[Ghost Of Highbury]

but i am sure that m1*m2 means perpendicular

[Ghost Of Highbury]

the third one is 0...

a=0 b=0 c=0

[nid404]

ooohh1!! nid,  chem practical?

yeah....why can't they name the damn things with the name of the substance rather than complicate it....FA1, FA2....F**k u...

and here's the m1m2=-1 proof

Slope of a given line= tan theta (delta y/ delta X)
So if there are two slopes m1 and m2
1 having inclination of theta, then m1=tan theta....Other line has inclination pi/2 + theta......i hope you can visualize this
then m2= tan (pi/2 +theta)=-cot theta(-1/tan theta) => m1m2=-1

[nid404]

 let p,with coorinates (p,q), be a fixed point on the 'curve ' with equation y=mx+c and let Q with coordinates (r,s), be any other point on y=mx+c . use the fact  thar the coordinates of p and Q satisfy the equation y=mx+c to  show that the gradient of PQ is m for all positions of Q.


ok..

Same logic...
m=tan theta
P=(p,q)
Q=(r,s)

tan theta= s-q / r-p
For any value of (r,s) on y=mx+c, (tan theta) will be constant, hence m will be constant

[astarmathsandphysics]

Maybe they meant to start from pre first principles ie the principle before the principle

[nid404]

Maybe they meant to start from pre first principles ie the principle before the principle

didn't get that