Using the coefficients to decide

The ratio test

You do not have to draw the lines to know how they sit. Write both equations in standard form and compare three ratios.

For $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$:

1. If $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$, the lines cross: exactly one solution, a consistent pair.
2. If $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$, the lines are parallel: no solution, an inconsistent pair.
3. If $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$, the lines coincide: infinitely many solutions, a dependent pair.

Why it works. A line $ax + by + c = 0$ with $b \neq 0$ has slope $-\dfrac{a}{b}$. Two lines have the same slope when $\dfrac{a_1}{b_1} = \dfrac{a_2}{b_2}$, and that is the same as $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2}$. Equal slopes mean parallel or coincident. The third ratio tells you which.

Three worked examples

1. Classify each pair.

• $3x + 2y - 8 = 0$ and $6x + 4y - 16 = 0$. The ratios are $\dfrac{3}{6}$, $\dfrac{2}{4}$ and $\dfrac{-8}{-16}$, all equal to $\dfrac{1}{2}$. The lines coincide.
• $x - 3y - 5 = 0$ and $2x - 6y - 7 = 0$. The ratios are $\dfrac{1}{2}$, $\dfrac{-3}{-6} = \dfrac{1}{2}$ and $\dfrac{-5}{-7} = \dfrac{5}{7}$. The first two are equal and the third differs. The lines are parallel.
• $4x + y - 6 = 0$ and $3x + 2y - 5 = 0$. The first two ratios are $\dfrac{4}{3}$ and $\dfrac{1}{2}$. They differ, so the lines cross.

2. Find $k$. Take the pair $x + 2y - 5 = 0$ and $3x + ky - 12 = 0$.

1. Compare the first two ratios. $\dfrac{a_1}{a_2} = \dfrac{1}{3}$ and $\dfrac{b_1}{b_2} = \dfrac{2}{k}$.
2. One solution. The ratios must differ: $\dfrac{1}{3} \neq \dfrac{2}{k}$, so $k \neq 6$.
3. No solution. The first two ratios are equal when $k = 6$. The third ratio is $\dfrac{5}{12}$, which differs from $\dfrac{1}{3}$, so $k = 6$ gives parallel lines.
4. Infinitely many. For this pair no value of $k$ works: the third ratio cannot be $\dfrac{1}{3}$. Change $12$ to $15$ and $k = 6$ gives the same line.

3. A real situation. Gym P charges ₹500 to join and ₹200 a month. Gym Q charges ₹200 to join and ₹300 a month. With $x$ months and $y$ rupees, the lines are $y = 500 + 200x$ and $y = 200 + 300x$. They meet at $(3, 1100)$. After $3$ months both gyms have cost ₹1100. For less than $3$ months Q is cheaper. For more, P is cheaper.

Practise with new numbers

Try it in Python

The program uses the ratio test to classify pairs. It uses fractions, so the ratios are exact.

Common mistakes

• Comparing the wrong ratios. Compare $\dfrac{a_1}{a_2}$ with $\dfrac{b_1}{b_2}$. Do not compare $\dfrac{a_1}{b_1}$ with $\dfrac{a_2}{b_2}$ in the formula.
• Forgetting the standard form. Move everything to one side first, because the sign of $c$ changes the third ratio.
• Stopping at two ratios. Equal first two ratios mean parallel or coincident. Check the third.
• Dividing by zero. If a coefficient is $0$, cross-multiply, or reason about the lines directly.

Remember these

• Different $\dfrac{a_1}{a_2}$ and $\dfrac{b_1}{b_2}$: one solution.
• Equal first two ratios, a different third: no solution.
• Three equal ratios: infinitely many solutions.
• To find $k$, set the ratios equal, then check the third ratio.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Classify $2x + 3y - 9 = 0$ and $4x + 6y - 18 = 0$.

2. Classify $x + y - 3 = 0$ and $2x + 2y - 5 = 0$.

3. Classify $x - y - 2 = 0$ and $2x + 3y - 9 = 0$.

4. Find $k$ so that $2x + 3y - 5 = 0$ and $4x + ky - 7 = 0$ have no solution.

5. Why must both equations be in the form $ax + by + c = 0$ before you use the test?