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# =S stuck on this maths questionnn. watch

1. AS MATHS - COORDINATE GEOMETRY
The point A has coordinates (7,4). The straight lines with equations x+3y+1=0 and 2x+5y=0 intersect at the point B. Find the coordinates of B and hence show that one of the two lines is perpendicular to AB.

I solved point B as (5,-2) simultaeneously. Now i'm stuck.

Rep to the first one that makes sense!
2. Find the equation of the line from the B coord. Then Check to see which one it is perpendicular to. New Gradient will be -1/oldgradient
3. (Original post by FranticMind)
Find the equation of the line from the B coord. Then Check to see which one it is perpendicular to. New Gradient will be -1/oldgradient
I tried doing this.
I assumed that coordinate A(7,4) and coordinate B(5,-2) joined and found the gradient of the line AB to be 3.
Do i re-arrange any one of the 2 formulae to find the gradient?
If so, what is the formula linking m1 and m2?
4. (Original post by FranticMind)
Find the equation of the line from the B coord. Then Check to see which one it is perpendicular to. New Gradient will be -1/oldgradient
oh okay so after that I tried doing this:
I assumed that coordinate A(7,4) and coordinate B(5,-2) joined and found the gradient of the line AB to be 3.
Do i re-arrange any one of the 2 equations to find the gradient?
If so, what is the formula linking m1 and m2?
5. kay so you found the gradient of AB to be 3, then pick one of the other equations, ie x + 3y + 1 = 0

rearrange to y = -1/3 x - 1/3

and you can see that the gradient is -1/3.

the gradient of a line perpendicular to AB would be -1/m > -1/3, and you've shown that it does have that gradient.
6. Find the gradient for line AB

Rearrange both the equations you have to get them in the form y=mx+c

Compare gradient for line AB to the equations.
7. Any two perpendicular lines will have product gradient -1, i.e. m1 x m2 = -1

So if a line has a gradient say -4, then any line perpendicular to this will have gradient 1/4. This is the same as flipping round the fraction (think of 4 as 4/1) and sticking a minus sign in front
8. (Original post by BrightGirl)
kay so you found the gradient of AB to be 3, then pick one of the other equations, ie x + 3y + 1 = 0

rearrange to y = -1/3 x - 1/3

and you can see that the gradient is -1/3.

the gradient of a line perpendicular to AB would be -1/m > -1/3, and you've shown that it does have that gradient.
omg thanks so much.
+rep
9. hehe okay GOT IT.
I owe rep to Goldfishy, sexuali, Brightgirl and FranticMind :P
10. (Original post by Goldfishy)
Any two perpendicular lines will have product gradient -1, i.e. m1 x m2 = -1

So if a line has a gradient say -4, then any line perpendicular to this will have gradient 1/4. This is the same as flipping round the fraction (think of 4 as 4/1) and sticking a minus sign in front
thaaankyouu
11. (Original post by Glee)
omg thanks so much.
+rep
you're welcome, don't worry about the rep
12. (Original post by sexuali)
Find the gradient for line AB

Rearrange both the equations you have to get them in the form y=mx+c

Compare gradient for line AB to the equations.
thankss
13. The perpendicular gradient will simply be 1/m and of inverse sign.

The gradient of the line AB is

(4- -2)/(7-5) = 3

The gradient of a perpendicular would be

-1/3

And x + 3y +1 =0

rearranges to y = (-1-x)/3

which is the same as y = -1/3(x+1) or y = -1/3x-1/3

thus the gradient is -1/3 and so it is the perpendicular to the first equation.

Hope that helped!
14. (Original post by BrightGirl)
you're welcome, don't worry about the rep
nahh i insist
15. oops I see brightgirl beat me to it. Whilst i was typing lol
16. (Original post by FranticMind)
The perpendicular gradient will simply be 1/m and of inverse sign.

The gradient of the line AB is

(4- -2)/(7-5) = 3

The gradient of a perpendicular would be

-1/3

And x + 3y +1 =0

rearranges to y = (-1-x)/3

which is the same as y = -1/3(x+1) or y = -1/3x-1/3

thus the gradient is -1/3 and so it is the perpendicular to the first equation.

Hope that helped!

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