# What is a and b if ax+by=0 passes through the points (9,8) and (8,9)

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The rule of the line that is passing through 2 given points is that the coordinates of the points are verifying the equation of the line.

Let's see how:

If the point (9,8) is on the line ax+by=0, then it's coordinates verify the equation of the line.

a*9 + b*8 = 0 (1)

Also, if the point (8,9) is on the line ax+by=0, then it's coordinates verify the equation of the line.

a*8 + b*9 = 0 (2)

Now, to determine a and b, we'll solve the system formed from the equations (1) and (2).

From (1), we'll get:

9a = -8b

We'll divide by 9 both sides:

a = -8b/9

We'll substitute a in (2):

-64b/9 + 9b = 0

-64b + 81b = 0

17b = 0

b = 0

a = 0

We are given that ax + by =0 passes through (9,8) and (8,9).

Therefore if we substitute the values of x and y in ax + by =0 we get the two equations:

a*9 + b*8 =0...(1)

a*8 + b*9 = 0...(2)

Subtract (2) from (1)

a - b =0

=> a=b

Putting this is (1)

9*a + 8*a =0

=> 17 a =0

=> a=0

**Therefore a = b =0. Or no straight line passes through (8, 9) and (9, 8).**

ax+by = 0 passes through (9,8) and (8,9).

Therefore the coordinates of the points (9,8) and (8,9) should satisfy the equation ax+by = 0.

(9,8) :

a*9+b*8 = 0...........(1)

(8,9):

a*8+b*9 = 0...........(2)

Add the equations (1) and (2): 17(a+b) = 0 Or a+b = 0

Also eq(2) - eq(1) gives: a-b = 0.

So

a+b =0 and

a-b = 0.

Adding 2a = 0. Or a = 0

Subracting 2b = 0. Or b =0.

So there is no line like ax+by = 0 passes through (9,8) and (8,9).

The line passing through Two points:

The line passing through 2 points (x1,y1) and (x2,y2) is :

y-y1 = {(y2-y1)/(x2-x1)}(x-x1). Put (x1,y1) = (9,8) and (x2,y2)= (8,9).

y-8 = {(9-8)/(8-9)}(x-9)

y-8 = -1(x-9)

x+y = 1.

Therefore the line passing through (9,8) and (8,9) is :

x+y = 1 making intercepts 1 and 1 on x and y axis respectively and this line cannot be reduced to the form ax+by = 0 which is a line through origin (making zero intercepts on x and y axis).