# Find the values of "a" for the line y= ax - 8 if it intercepts the x axis at the point where x=2.

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### 3 Answers

We have the x-intercept of the line y = ax - 8 as x = 2.

Now we can rewrite y = ax - 8

y = ax - 8

=> ax - y =8

=> ax/8 - y/8 = 1

This is of the form x/a + y/b= 1, where a and b are the intercepts.

Therefore 8/a = 2

=> a = 4

**Therefore we get the value of a as 4.**

Given the line y = ax-8

Also, given that the line (y) intercepts with the x-axis at the point x=2.

Then the values of y when x =2 is 0.

Then, we know that the point (2,0) is on the line y.

Let us substitute.

==> y= ax-8

==> 0 = 2a -8

==> 2a = 8

**==> a = 4**

**==> y= 4x -8**

**Then, the values of a is 4.**

y = ax-8 intersects x axis at x = 2.

We know the equation of x axis is y = 0.

Therefore y= ax-8 ...(1) and y = 0...(2) are the two lines whose intersection is x=2, and y= 0. Or at (x,y) = (2,0).

So we put y = 0 in (1) and get:

0 = ax-8.

Therefore ax= 8, or a = 8/x. but x = 2,

So a = 8/2 = 4.

Therefore the value of a = 4 in order that y = ax-8 and y = 0 intersect at x = 2.