HomeReading the Constant of Proportionality from Graphs Free original lessons
Reading the Constant of Proportionality from Graphs
Learn to extract k from proportional lines by computing y/x at marked points, writing y = kx, and checking with slope triangles.
Free lesson
Four exercises
Worked answers
Algebra and Function Practice
A proportional relationship graphs as a straight line through the origin. The constant of proportionality k is the ratio y/x for any point on the line (except the origin). This lesson shows how to read k from plotted lines, write the equation y = kx, and verify your value using a slope triangle. You will practice with marked points and simple arithmetic, using consistent units and clearly labeled axes.
Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.
Identify a proportional line and its marked point
A proportional relationship has a graph that is a straight line passing through the origin (0,0). If the line does not pass through the origin, it is not proportional, and k is not defined in this lesson.
Each graph plate shows one line with at least one marked point (x, y). The coordinates are read from the axes, which have consistent units. Choose a marked point other than the origin to compute k.
- Confirm the line passes through (0,0).
- Locate a marked point and read its x and y values from the axes.
- Record the point as an ordered pair (x, y) with units noted.
Compute k = y/x and write the equation
For a proportional line, k equals y divided by x for any point on the line except the origin. Use the marked point to compute k. Keep units consistent; if x is in hours and y is in dollars, k is dollars per hour.
After computing k, write the equation in the form y = kx. Substitute the marked point into the equation to check that both sides match.
- Divide the y-coordinate by the x-coordinate: k = y/x.
- Simplify the fraction or decimal using simple arithmetic.
- Write y = kx with your value of k.
- Substitute the marked point to verify the equation holds.
Verify with a slope triangle
A slope triangle uses two points on the line to show rise over run. For a proportional line, rise/run equals k. Draw a right triangle with horizontal leg run = x2 − x1 and vertical leg rise = y2 − y1. The ratio rise/run should match y/x from any point.
Use the origin and the marked point for a simple triangle: run = x, rise = y, so rise/run = y/x = k. This confirms your value without relying on a single calculation.
- Choose two points on the line, such as (0,0) and the marked point.
- Compute run = difference in x and rise = difference in y.
- Divide rise by run and compare with your k.
- If they match, the equation y = kx is consistent with the graph.
Handle fractions, decimals, and unit labels
k may be a fraction or decimal. Keep the exact fraction if possible, then convert to a decimal only if needed. Label k with units derived from the axes, such as meters per second or cost per item.
When axes use different scales, read coordinates carefully. The value of k does not depend on the scale as long as you read both coordinates correctly. Always state units in your final answer.
- Simplify the fraction y/x by dividing numerator and denominator by common factors.
- If using decimals, round only at the final step and state the rounding.
- Attach units to k by dividing the y-axis unit by the x-axis unit.
- Write the equation with k and note the units in context.
Worked example
A graph shows a straight line through the origin with a marked point at (4, 10). The x-axis is in hours and the y-axis is in dollars. Find k, write the equation, and verify with a slope triangle.
Worked answer
k = 10/4 = 2.5 dollars per hour. Equation: y = 2.5x. Slope triangle from (0,0) to (4,10): run = 4, rise = 10, rise/run = 10/4 = 2.5, which matches k.
Four practice exercises
Exercise 1
A proportional line passes through the origin and has a marked point at (3, 12). The x-axis is in seconds and the y-axis is in meters. Find k and write the equation.
Show answer and check
k = 12/3 = 4 meters per second. Equation: y = 4x.
Why: For a proportional line, k = y/x. Using (3,12), k = 12/3 = 4. The equation is y = 4x with units meters per second.
Check: Substitute (3,12) into y = 4x: 12 = 4(3) = 12. Also draw a slope triangle from (0,0) to (3,12): rise 12, run 3, ratio 4.
Exercise 2
A graph shows a line through the origin with a marked point at (5, 7). The axes have no units. Compute k as a fraction and as a decimal rounded to two places. Write the equation.
Show answer and check
k = 7/5 = 1.4. Equation: y = 1.4x (or y = (7/5)x).
Why: k = y/x = 7/5. The fraction simplifies to 7/5, which equals 1.4. The equation uses this constant.
Check: Substitute (5,7): 7 = 1.4(5) = 7. Slope triangle from origin: rise 7, run 5, ratio 7/5 = 1.4.
Exercise 3
A proportional line through the origin has a marked point at (2, 9). The x-axis is in items and the y-axis is in dollars. Find k, write the equation, and state the unit of k.
Show answer and check
k = 9/2 = 4.5 dollars per item. Equation: y = 4.5x.
Why: k = y/x = 9/2 = 4.5. Units are dollars divided by items, so dollars per item. The equation is y = 4.5x.
Check: Substitute (2,9): 9 = 4.5(2) = 9. Slope triangle: rise 9, run 2, ratio 4.5.
Exercise 4
A line through the origin has a marked point at (8, 6). Another point on the line is (4, 3). Compute k using both points and confirm they agree. Write the equation.
Show answer and check
Using (8,6): k = 6/8 = 3/4 = 0.75. Using (4,3): k = 3/4 = 0.75. Equation: y = 0.75x (or y = (3/4)x).
Why: Both points give the same ratio y/x = 3/4. Since the line passes through the origin, k is constant. The equation uses k = 3/4.
Check: Substitute each point into y = 0.75x: 6 = 0.75(8) = 6 and 3 = 0.75(4) = 3. Slope triangle between points: run = 8−4 = 4, rise = 6−3 = 3, ratio 3/4 = 0.75.
Common mistakes
- Using a point that is not on the line or misreading coordinates from the axes. Trace the line to the marked point and read both coordinates along grid lines. Confirm the point satisfies y = kx after computing k.
- Dividing x by y instead of y by x. Remember k = y/x. Check by substituting the point into y = kx; if it fails, you likely inverted the ratio.
- Assuming any straight line is proportional. Verify the line passes through the origin. If it does not, k is not defined for a proportional relationship in this lesson.
- Ignoring units or mixing units on the axes. Write units for each axis and compute k with matching units. State k as y-unit per x-unit.
Self-check rubric
- Correctly identifies the line as proportional by checking it passes through the origin.
- Reads the marked point coordinates accurately from the axes.
- Computes k = y/x with correct arithmetic and consistent units.
- Writes the equation y = kx using the computed k.
- Verifies k using a slope triangle or substitution.
- States units for k and explains the meaning in context.
- Uses clear notation and shows steps; accepts equivalent fractions or decimals.
Questions and answers
Can k be negative?
Yes, if the line passes through the origin and has a negative slope, k is negative. Compute y/x using a marked point with a negative coordinate.
What if the marked point has x = 0?
The origin gives 0/0, which is undefined. Choose a different marked point with x not equal to zero.
How do I check my k without another point?
Substitute the marked point into y = kx. If both sides are equal, your k is consistent with that point. You can also use a slope triangle from the origin to the point.
Do I need to simplify fractions?
Simplifying helps compare values and write cleaner equations. Keep the exact value; decimals are acceptable if stated clearly.