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

Answer:

The greatest possible value for b is 26.

Step-by-step 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: $\frac{y_2 - y_1}{x_2 - x_1}$   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 = $\frac{b - 14}{-2}$

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:  $\frac{72}{14 - b }$

Since, the slopes are equal we can equate:

$\frac{b - 14}{-2} = \frac{72}{14 - b}$

$\implies \frac{b - 14}{-2} = \frac{72}{-(b - 14)}$

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

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

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.