0580 · 3.7Perpendicular lines

Negative reciprocal gradients

0:00Two straight lines can meet at many different angles, but perpendicular lines meet at a right angle. Their directions are linked in a very precise way. Once we understand that link, we can calculate a perpendicular gradient without drawing the lines. We can also find a perpendicular line through a particular point, including the perpendicular bisector of a segment. We need two familiar ideas about straight lines. Gradient means change in y divided by change in x.

0:24In the equation y equals m times x plus c, m is the gradient and c is the y intercept. The gradient controls the direction of a line, while the intercept controls where it crosses the vertical axis. A positive gradient means the line rises as we move from left to right. A negative gradient means it falls in that direction. For example, a gradient of 3/2 means moving two units right corresponds to moving three units up.

0:49Thinking about these movements will help us understand perpendicular gradients, rather than just remembering a rule. Imagine a line through the origin with gradient 3/2. One movement along it takes us two units right and three units up. Rotate that movement clockwise through ninety degrees. The original rightward movement becomes downward, while the original upward movement becomes rightward. The rotated movement therefore takes us three units right and two units down.

1:14The perpendicular line has a vertical change of negative two and a horizontal change of three. Its gradient is therefore −2/3. The original movement lengths have swapped places, and one direction has reversed. That explains both parts of the perpendicular gradient rule. We reverse the fraction and change its sign. Reversing a fraction gives its reciprocal, so this result is called the negative reciprocal. 3/2 becomes −2/3. Multiplying those gradients gives negative one because the matching factors cancel.

1:45For perpendicular lines whose gradients are defined, the gradients multiply to negative one. This is another way of expressing the same relationship. The sign change alone is not enough. A gradient of two has perpendicular gradient −1/2, not negative two. Likewise, −3/4 becomes positive 4/3. A steep rising line becomes a shallow falling line after a right-angle turn. The negative reciprocal captures both the change in direction and the change in steepness.

2:14Our first exam-style example asks us to find the gradient of a line perpendicular to 2y = 3x + 1. First, divide every term by two. The equation becomes y = 3/2 × x, plus one half. The given line therefore has gradient 3/2. Take the negative reciprocal of 3/2. This gives −2/3 as the perpendicular gradient. Check by multiplying 3/2 by −2/3. The product is negative one, confirming the required relationship.

2:44Notice that the intercept does not enter this calculation because it affects position, not direction. A common mistake here is to read the gradient as three from the original equation. That coefficient only gives the gradient when y has coefficient one. Rearranging first prevents this error. Another mistake is to change the sign without reversing the fraction. Use both parts of the negative reciprocal process, then check the product.

Equation through a given point

3:08A gradient alone does not identify one particular line. Many lines can share the same direction but cross the vertical axis in different places. To find an equation, we also need a point that the required line passes through. Substituting that point into y equals m times x plus c allows us to calculate the missing intercept. Our second example asks us to find the equation of a line perpendicular to y = −2x + 7.

3:33The required line passes through the point with x equal to four and y equal to one. The given gradient is negative two, which we can write as −2/1. Its negative reciprocal is positive 1/2. Write the required equation as y = ½x + c. Substitute the coordinates to obtain 1 = ½ of four, plus c. Simplifying gives 1 = 2 + c. Subtract two from both sides to find c = −1.

3:59The equation is y = ½x − 1. We can check the answer in two separate ways. Multiplying negative two by 1/2 gives negative one, so the directions are perpendicular. Substituting x equal to four into our answer gives 2 − 1, which equals one. The line therefore passes through the required point as well. Both conditions matter when finding its equation.

Finding a perpendicular bisector

4:21A perpendicular bisector combines two conditions in its name. Perpendicular means it meets the segment at a right angle. Bisector means it divides that segment into two equal lengths. It must therefore pass through the midpoint and have a gradient perpendicular to the segment. A right-angle crossing somewhere else is not a perpendicular bisector. Our third example uses the segment joining two given points. Point A is negative three comma eight, and point B is nine comma negative two.

4:49We first calculate the midpoint because the perpendicular bisector must pass through it. For the midpoint’s x coordinate, add negative three and nine to obtain six. Divide six by two to get three. For its y coordinate, 8 + −2 also gives six. Halving that result gives three, so the midpoint has x equal to three and y equal to three. Calculate the segment’s gradient using the same point order in both coordinate differences.

5:15−2 − 8 gives a vertical change of negative ten. Across, 9 − −3 gives twelve. Dividing produces −10/12, which simplifies to −5/6. Its negative reciprocal is positive 6/5. The bisector’s equation begins as y = 6/5 × x, plus c. Substitute the midpoint to obtain 3 = 6/5, times three, plus c. Simplifying gives 3 = 18/5 + c. Subtract 18/5, writing three as 15/5. The intercept is therefore −3/5.

5:52Our answer is y = 6/5 × x, minus 3/5. Its gradient is perpendicular to the segment’s gradient because their product is negative one. It also passes through the midpoint because we used that point to calculate the intercept. These two facts establish that it is the perpendicular bisector, rather than merely another perpendicular line. In an examination, show the midpoint, the segment’s gradient, the perpendicular gradient and the substitution for the intercept.

6:20Keep coordinate subtraction in the same order throughout the gradient calculation. Using an endpoint instead of the midpoint produces a perpendicular through that endpoint. It does not bisect the segment, so it answers a different question.

Horizontal and vertical lines

6:32Horizontal and vertical lines need separate treatment. A horizontal line has gradient zero, while a vertical line has an undefined gradient because its horizontal change is zero. They are perpendicular to each other, but we cannot apply the negative reciprocal calculation to zero. For example, the perpendicular to a horizontal line through our previous midpoint has equation x = 3. These methods must work without a calculator on Paper two, and they also apply on calculator Paper four.

6:59Keep fractions exact in these examples rather than replacing them with rounded decimals. For answers requiring approximation, the usual instruction is three significant figures unless stated otherwise. Do not round during intermediate calculations because that can change the final result. Perpendicular lines meet at a right angle, and their defined gradients multiply to negative one. The negative reciprocal comes from rotating a movement through ninety degrees, which swaps its horizontal and vertical parts.

7:26To find an equation, combine that gradient with a point on the required line. For a perpendicular bisector, that point must be the segment’s midpoint.