Are the points , and on the same straight line? Suggest a method to check this without plotting and joining the points.

Question

Homi · Accepted Answer

Understand the Question

To determine whether three points $M(-3, -4)$ $M (- 3, - 4)$ , $A(0, 0)$ $A (0, 0)$ , and $G(6, 8)$ $G (6, 8)$ lie on the same straight line (are collinear) without plotting them, we can use an algebraic method:
- Slope Method: If the slope of line segment $MA$ equals the slope of line segment $AG$ , the points lie on the same line since $A$ is a common point.
- Distance Formula Method: If the sum of the lengths of the two shorter segments equals the length of the longest segment ( $MA + AG = MG$ ), then the points are collinear.

Step 1 · Calculate the Slope of MA and AG

The slope $m$ of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

m = \dfrac{y_2 - y_1}{x_2 - x_1}

Calculate the slope of segment $MA$ with $M(-3, -4)$ and $A(0, 0)$ :

\begin{aligned} \text{Slope of } MA &= \dfrac{0 - (-4)}{0 - (-3)} \\[0.6em] &= \dfrac{4}{3} \end{aligned}

Calculate the slope of segment $AG$ with $A(0, 0)$ and $G(6, 8)$ :

\begin{aligned} \text{Slope of } AG &= \dfrac{8 - 0}{6 - 0} \\[0.6em] &= \dfrac{8}{6} \\[0.6em] &= \dfrac{4}{3} \end{aligned}

Step 2 · Verify Collinearity

Since

\text{Slope of } MA = \text{Slope of } AG = \dfrac{4}{3}

and point $A$ is common to both segments, the points $M$ , $A$ , and $G$ lie on the same straight line.

Answer

Yes, the points $M(-3, -4)$ , $A(0, 0)$ , and $G(6, 8)$ lie on the same straight line.

Common Mistakes

Missing Common Point Justification: Equal slopes only guarantee parallel lines; pointing out that $A$ is a common point is necessary to prove they lie on the exact same line.
Sign Errors in Slope Formula: Miscalculating $\dfrac{0 - (-4)}{0 - (-3)}$ as negative by dropping double negatives.
Incorrect Coordinate Order: Calculating $\dfrac{x_2 - x_1}{y_2 - y_1}$ (run over rise) instead of $\dfrac{y_2 - y_1}{x_2 - x_1}$ (rise over run).

Question 6