Question

A line passes through the origin and through points A(−2, b−14) and B(14−b, 72). What is the greatest possible value of b?

Answer 1

The greatest possible value for b is 26.

Explanation:

Given that the line passes through the Origin O(0, 0); A(-2, b - 14) &

B(14 - b, 72).

Let us assume the points are in the order: AOB.

Since the line passes through all these points the slope of the line segment AO = The slope of the line segment AB.

Slope of a line with two points:
`$ (y_2 - y_1)/(x_2 - x_1) $` where
`$ (x_1, y_1) $` and
`$ (x_2, y_2) $` are the points given.

`$ (x_1, y_1) = (0,0) $`

`$ (x_2, y_2) = (-2, b - 14) $`

Therefore, the slope of the line segment AO =
`$ (b - 14)/(-2) $`

Similarly, for the slope of the line segment OB.

The two points are
`$ (x_1, y_1) = (0, 0) $` and
`$ (x_2, y_2) = (14 - b, 72) $`.

The slope is:
`$ (72)/(14 - b ) $`

Since, the slopes are equal we can equate:

`$ (b - 14)/(-2) = (72)/(14 - b) $`

`$ \implies (b - 14)/(-2) = (72)/(-(b - 14)) $`

`$ \implies (b - 14)^2 = 72 * 2 = 144 $`

`$ \implies (b - 14)^2 = 12^2 $`

Taking square root on both sides we get:

`$ \implies (b - 14) = \pm 12 $`

`$ \implies b = 2 \hspace{2mm} or \hspace{2mm} 26 $`

Therefore, the maximum value of b = 26.

Hence, the answer.