CEMC Cayley, 2015

Algebra Difficulty 1.4 Junior Find the answer Canada

The points $(-1, q)$ and $(-3, r)$ are on a line parallel to $y = \tfrac{3}{2}x + 1$ . What is the value of $r - q$ ?

Pick one

Solution

The line with equation $y = \frac{3}{2}x+1$ has slope $\frac{3}{2}$ .

Since the line segment joining $(-1,q)$ and $(-3,r)$ is parallel to the line with equation $y = \frac{3}{2}x+1$ , then the slope of this line segment is $\frac{3}{2}$ .

Therefore, $\dfrac{r-q}{(-3)-(-1)} = \dfrac{3}{2}$ or $\dfrac{r-q}{-2} = \dfrac{3}{2}$ .

Thus, $r-q = (-2)\cdot\dfrac{3}{2}=-3$ .

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