0580 · 3.6Parallel lines

Gradients and parallel lines

0:00Two straight lines can cross the vertical axis at different heights while pointing in exactly the same direction. Their positions differ, but their steepness matches. That connection lets us find a parallel line without drawing an accurate graph. By the end, you will understand how its gradient and a single point determine its equation. Parallel lines are straight lines in the same plane that never meet, however far they are extended.

0:22On a coordinate grid, distinct lines with the same gradient are parallel, provided neither is vertical. Gradient measures the change in y divided by the corresponding change in x. It describes how steeply a line rises or falls. For example, a gradient of two means y increases by two whenever x increases by one. Both lines on this grid make that same movement. Starting at different heights does not change their direction.

0:48Their vertical separation remains unchanged as we move across the grid, so neither catches up with the other. We will use the equation y equals m times x plus c. Here, m is the gradient, and c is the y-intercept. The y-intercept tells us where the line crosses the vertical axis, where x = 0. For a parallel line, we keep the gradient and find the intercept using the given point.

1:10This gives us two separate pieces of information. Being parallel fixes the direction, but many lines can have that direction. Passing through a particular point fixes the position. Substituting that point works because its x and y values must satisfy the equation of any line passing through it.

Finding equations from given lines

1:27Find the equation of the line parallel to y = 4x − 1. It must pass through the point one comma negative three. The given gradient is four. Our parallel line therefore has gradient four as well. Write its equation as y = 4x + c, leaving the intercept unknown. Substituting the point gives negative three equals four times one plus c. Multiplying gives −3 = 4 + c. Subtracting four gives c = −3 − 4.

1:56Therefore, c = −7. The required equation is y = 4x − 7. Check both conditions before finishing. Substituting x equal to one gives 4 − 7, which is negative three, so the line passes through the point. Its gradient is four, matching the original line. The intercept changed from negative one to negative seven because we needed a different position, not a different direction. A common mistake is copying the original intercept as well as the gradient.

2:24That simply reproduces the original line, which may not pass through the required point. Another mistake is treating the point’s y-coordinate as the intercept. That only works when the point lies on the vertical axis, where x = 0. Find the gradient and equation of the line parallel to 2x + 3y = 6 through the point three comma one. First, subtract 2x to obtain 3y = −2x + 6.

2:49Dividing every term by three gives y = −⅔x + 2. The parallel gradient is therefore negative two thirds. Write the new equation as y = −⅔x + c. Substitution gives one equals negative two thirds times three plus c. Simplifying gives 1 = −2 + c. Adding two gives c = 3. The required equation is y = −⅔x + 3. The coefficient of x gives the gradient only after y has been isolated with coefficient one.

3:23Reading two as the gradient from the original equation would therefore be wrong. Keep the negative sign because this line slopes downwards from left to right. Parallel lines match both steepness and direction, not just the size of the gradient.

Using points and special cases

3:36An exam question may give two points instead of an equation. Suppose the original line passes through negative one comma two and three comma four. Find a parallel line through two comma negative three. The change in y is 4 − 2, giving two. Across, 3 − −1 gives four. Dividing two by four gives a gradient of one half. The parallel equation is y = ½x + c. Substitution gives −3 = ½ of two, plus c.

4:06Simplifying gives −3 = 1 + c. Subtracting one gives c equal to negative four, so the equation is y = ½x − 4. These methods must work without a calculator on Paper Two and remain useful on calculator Paper Four. Show the gradient, the substitution and the final equation clearly. Keep fractions exact here rather than rounding them to decimals. There is no need to round these exact equations to three significant figures.

4:34Horizontal lines fit the same idea because their gradient is zero. Vertical lines are the exception because their gradient is undefined. Parallel vertical lines have equations such as x = 2 and x = 5. Also, equal gradients and equal intercepts describe the same line, not two distinct parallel lines. Matching direction alone does not guarantee different positions. The essential connection is that parallel lines that are not vertical have the same gradient.

5:01Use that gradient to write the new equation with an unknown intercept. Substitute the given point to determine its position. Finally, check that your equation has the correct gradient and contains the required point.