# The gradient of the line joining the points (9,a) & (2a,1) is 2/a, where "a" Is not equal to "0", calculate the possible values of "a".

*print*Print*list*Cite

We have the points: (9,a) and (2a, 1)

The gradient of the line m = 2/a (a is not 0)

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

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

Cross multiply:

==> a(1-a) = 2(2a-9)

Open brackets:

==? a - a^2 = 4a - 18

Now group similar:

==> a^2 + 3a - 18 = 0

Now factor:

==> (a + 6)(a-3) = 0

==> a1= -6

==> a2= 3

The gradient of the segment is the slope of that segment.

We know that the slope of a line that passes through 2 given points is:

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

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

We know, from enunciation , that m = 2/a, so we'll substitute my by it's given value.

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

We'll cross multiply:

2(2a-9) = a(1-a)

We'll remove the brackets:

4a - 18 = a - a^2

We'll move all terms to one side and combine like terms:

a^2 + 3a - 18 = 0

We'll apply the quadratic formula:

a1 = [-3 + sqrt(9 + 72)]/2

a1 = (-3+9)/2

a1 = 6/2

a1 = 3

a2 = (-3-9)/2

a2 = -6

**The values of a are: {-6 ; 3}**

The gradient m of the line joining A and B is given by:

m = (yB-yA)/(xB-xA)............(1)

A(9,a) and B(2a,1). m =1/a. Substitute the of m and coordinates of A and B in the relation at (1)

(1-a)/((2a-9) = 2/a. Multiply by the denominators' LCM, (2a-9)a.

a(1-a) = 2(2a-9)

a-a^2 = 4a-18

a-a^2-4a+18 = 0. Multiply by -1.

a^2 -3a-18 = 0

a^2-3a-18 = 0

a^2-6a +3a-18 = 0

a(a-6) +3(a-6) = 0

(a-6)(a+3) = 0

a-6 = 0 or a+3 = 0

a=6 or a=-3.