Maths (simultanius equations) ?

Question

The Point With CO-ordinates (1,3) and (3,9/2) lies on the line with the equation

ax-by = 2

a)find the value of 'a' and 'b' ?
b)Find the Gradiant of the line ?

Answer 1

the answer goes like this.
substitute the points.
i.e x---x co-ordinate of the point and y--y co_ordinate of the points.
so it turns out to be ...
x-3y=2 and
3x-(9/2)y= 2
and so ur equations will be x-3y =2
and 6x-9y =4
now try gauss elimination or substitution method.or elimination..
and ur answer comes to be x= -2/3
and y= -8/9
and as for the gradient
the formula is m=y2-y1/(x2-x1)( it is also called as slope)
and so ur gradient will be m=3/4
got it?

Answer 2

The equation has to be true for any point on the line. So it's true for both (1,3) and (3,9/2)
ax-by = 2
Substituting the first point...
a(1) - b(3)=2 or
(1) a-3b=2

Substituting the second point...
a(3) - b(9/2) = 2
multiplying both sides by 2 to get rid of that fraction...
(2) 6a - 9b = 4

So your simultaneous equations are
(1) a-3b=2
(2) 6a - 9b = 4

You can solve them by "Addition"/"Elimination"
You can solve them by substitution
You can solve them by determinants.
Or you can use some combination of these.

I'd do it for you, but then I'd be depriving you of the fun.

Once you get a and b, and have the equation
ax-by = 2, with values for a and b,
solve for y, you will have an equation that looks kind of like
y = mx + d.
The usual format of the equation is y = mx + b, but you're working with b as a variable already and I don't want to confuse you.
Anyway, m is the slope of the line.

If by "gradiant" you mean "slope," you're done. But if you mean "gradient" you have some more work to do.

Gradient refers to the slope of the line. But, rather than a mathematically useful number, it's expressed in % of 90 degrees. e.g., 45 degrees is 50% of 90 degrees, so the gradiant is 50%. Engineers use it... so do architects.

The problem now is to find the angle. You need to find the arctangent of the slope. If you know trigonometry, you'll understand, If you don't, it really doesn't make any difference. You'll need a calculator with a key ATAN or Tan⁻¹ key, or you'll need a spreadsheet. The MS Excel function is ATAN.

So, when you get the angle as indicated above, divide it by 90. That's the gradient.

Answer 3

[Part 1]

We have two points on the line: (1,3) and (3, 9/2)

We have an equation for the line: ax - by = 2

Now, just plug the point into the equation, forming two simultaneous equations:

1) a(1) - b(3) = 2 ----> a - 3b = 2

2) a(3) - b(9/2) = 2 ----> 3a - 9b/2 = 2 ----> 6a - 9b = 4

Now, try multiplying one of the equations by some factor that would allow you to cancel out one variable when adding the two equations. I'll multiply equation #1 by '-6,' and then i'll add that result to equation #2:

-6a + 18b = -12
6a - 9b = 4 (add these two equations)
----------------------------------------------------
9b = -8 -----> b = -8/9

Now let's plug 'b' into one of the original equations (i'll randomly pick the original form of equation #2):

3a - 9b/2 = 2

3a - (1/2)9(-8/9) = 2

3a + 4 = 2 -----> a = -2/3

*** a = -2/3 and b = -8/9

[Part 2]

What you call the gradient of the line, in American, we call it the slope, and we represent it by the letter 'm.'

m = (y2 - y1)/(x2 - x1)

m = (9/2 - 3) / (3 - 1)

m = (3/2) / 2

*** m = 3/4

Answer 4

Ok you have 2 equations

a(1) - b(3) =2
a(3) - b(9/2) = 2

a = 2+3b from first equation

3*(2+3b)-9/2 b = 2
6 + 9b - (9/2)b = 2
9/2 b = -4
b = -8/9

Then a = 2+3b = 2-24/9 = -6/18 = -1/3

Answer 5

Create two equations, each with one set of coordinates, such as
1a + 3b =2 and 3 a +9/2 b =2
They can be solved by conventional means for values of a and b.
The "gradient" of the line is found by converting the equation ax-by=c to the form y = -c/b+ax/b.
a/b is the slope or gradient.