Animated Solution for Mathematics - Straight Lines: The points $(0,8/3),(1,3)$ and $(82,30)$ are vertices of

We are given three points: $A(0,\frac{8}{3})$, $B(1,3)$, and $C(82,30)$. The objective is to identify the nature of the triangle formed by these vertices.

While the instinct to calculate side lengths using the distance formula is strong, this approach is often a trap in competitive examinations. We must first investigate whether these points actually form a triangle or if they are collinear.

If three points are collinear, they lie on a single straight line. In such a case, the area of the triangle they would "form" is exactly $0$.

To determine if these points are collinear, we do not need the distance formula. Instead, we use the concept of the slope, which represents the constant steepness of a line.

Your Most Elegant Weapon

The slope $m$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is defined by the ratio of the vertical rise to the horizontal run:

If the slope of segment $AB$ is equal to the slope of segment $BC$, the points $A$, $B$, and $C$ must lie on the same line.

A Dance of Numbers

First, we calculate the slope $m_1$ for segment $AB$:

Simplifying the numerator, we have $3-\frac{8}{3}=\frac{9}{3}-\frac{8}{3}=\frac{1}{3}$. Thus, the slope is:

Next, we calculate the slope $m_2$ for segment $BC$:

Reducing the fraction by dividing both the numerator and denominator by $27$, we obtain:

Since $m_1=m_2=\frac{1}{3}$, the slopes are identical. This confirms that the points $A$, $B$, and $C$ are perfectly collinear.

Because these points lie on the same straight line, they cannot form a triangle. Given the standard options for such a problem, the correct conclusion is none of these.