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This page sets out the method for parallel & perpendicular lines in the order you would use it, then walks through the full lesson explanation. For three complete worked examples with the slips that cost marks, see the Parallel & perpendicular lines topic page.
When you see 'parallel to' or 'perpendicular to' a line 'through the point (…)', find the new gradient first and the intercept second.
When you see 'perpendicular to' and a point the new line must go through.
Parallel lines have equal gradients. The intercept can change.
A line parallel to y=4x-2 through the point (0, 1) is y=4x+1. If the point is not on the y-axis, put it into y = mx + c. For gradient 4 through (2, 9): 9 = 4×2 + c, so c = 1 and y = 4x + 1.
Perpendicular to gradient 2 is -1/2. Flip the gradient and change its sign. The product of the two gradients is -1.
The gradient of a line parallel to y=4x is 4. You do not need a new point for that.
A solution of two line equations is a point that lies on both lines. Two distinct parallel lines never meet, so they have 0 solutions.
If the gradients differ, the lines meet once.
A line written as y=3x-1 has gradient 3 and crosses the y-axis at -1. A parallel line also has gradient 3. If it must pass through (2, 4), start from y-4=3(x-2), not from y=3x-1. Expanding gives y=3x-6+4, so y=3x-2. Check the point: 3 × 2-2=4. Check a second point on your line if you have time, but the given point is the one that must fit.
The equation 3x+4y=12 does not have gradient 3. Make y the subject: 4y=-3x+12, so y=-3/4x+3. The gradient is -3/4. The intercept is 3, which you can see by putting x=0. Students who read the coefficient of x straight from the original equation get every later gradient wrong, including the perpendicular one.
Method: 1. Rearrange into y=mx+c before you read a gradient. 2. Parallel lines have the same m. 3. Use the given point in y-y_1=m(x-x_1). 4. Put the point back into your finished line.
The negative reciprocal of 2 is -1/2, because 2 × (-1/2)=-1. The negative reciprocal of -3/4 is 4/3. Do both jobs: flip, and change the sign. Doing only one of them is the standard loss of a mark. A gradient of 0 is a horizontal line; it has no negative reciprocal in the same way, because a vertical line has an undefined gradient. You will not be asked to flip 0 in ordinary questions.
Through (4, 1), the line perpendicular to y=2x+5 has gradient -1/2. Then y-1=-1/2(x-4). Expand: y=-1/2x+2+1, so y=-1/2x+3. At x=4 the right-hand side is -2+3=1, which is the y-coordinate you started with. The original line has gradient 2, and 2 × (-1/2)=-1, so the directions really are perpendicular.
Method: 1. Write the known gradient as a fraction. 2. Turn it upside down and change the sign. 3. Check that the two gradients multiply to -1. 4. Then build the line through the given point.
Questions often want ax+by+c=0 with integer coefficients. Start from the point form so the fractions have a clear denominator. For gradient 4/3 through (1, -2): y+2=4/3(x-1). Multiply by 3: 3y+6=4x-4. Bring every term to one side: 0=4x-3y-10, so 4x-3y-10=0. You could multiply by -1 and write -4x+3y+10=0. Either is acceptable if it is equivalent and the point still fits.
Always substitute the given point back into the final equation. For (1, -2) in 4x-3y-10: 4(1)-3(-2)-10=4+6-10=0. If you get anything else, the intercept is wrong or a sign flipped when you cleared the fraction. Also check the gradient product against the original line. Here the original gradient was -3/4 and (-3/4)×4/3=-1. Two different checks, two different mistakes caught.
Method: 1. Clear the fraction by multiplying through by the denominator. 2. Collect every term on one side. 3. You may multiply by -1 so the first coefficient is positive. 4. Test the given point in the integer equation.
To compare two lines, put both into y=mx+c and look only at m. Take 2x+5y=10 and 5x-2y=8. The first becomes y=-2/5x+2. The second becomes y=5/2x-4. The product (-2/5)×5/2=-1, so the lines are perpendicular. They are not parallel, because the gradients are not equal. If the product had been 1 or 4 or any number other than -1, and the gradients differed, the answer would be neither.
A line crossing the x-axis has y=0. That is a substitution, not a new gradient. The line perpendicular to 2x-5y+10=0 has gradient -5/2, because the original gradient is 2/5. Through (5, 1), set y=0 in y-1=-5/2(x-5) to get x=27/5. Write the coordinates as a pair, (27/5, 0). A bare 5.4 with no y-coordinate is incomplete if the question asked for coordinates.
Method: 1. Find both gradients from the two equations. 2. Equal gradients mean parallel. 3. A product of -1 means perpendicular. 4. Anything else is neither. Say which, and show the test.
The rule for perpendicular lines is that the two gradients multiply to -1. Turn the gradient upside down and change its sign. A gradient of 3/4 gives -4/3. A gradient of -5 is -5/1, so the perpendicular gradient is 1/5. A gradient of 1/2 gives -2. Check each time by multiplying: 3/4×(-4/3)=-1. If the product is not -1, one of the two steps was missed.
Be careful when the equation is not in the form y=mx+c. The line 2x+3y=6 has 3y=-2x+6, so y=-2/3x+2 and its gradient is -2/3, not -3/2 and not 2. The perpendicular gradient is therefore 3/2. Do the rearrangement first, and only then flip and change the sign. Parallel lines keep the gradient exactly as it is, so for a parallel line you copy it, and for a perpendicular line you transform it.
Method: 1. Rearrange the given line into y=mx+c and read m. 2. For a perpendicular line, flip the fraction and change the sign. 3. Multiply the two gradients to confirm the product is -1. 4. Use the new gradient with the given point to find the intercept.
Tip: Write the pair of gradients next to each other and multiply them. It is a ten second check that shows up errors with signs.
The perpendicular bisector of AB cuts AB in half at a right angle. It needs two ingredients: the midpoint of AB, which is a point the new line passes through, and a gradient perpendicular to AB. For A(1, 2) and B(7, 6) the midpoint is the average of each coordinate, ((1+7)/2,(2+6)/2)=(4, 4). The gradient of AB is (6-2)/(7-1)=2/3, so the perpendicular gradient is -3/2.
Now use y=-3/2x+c with the point (4, 4). Substituting gives 4=-6+c, so c=10, and the line is y=-3/2x+10. A common slip is to use A or B instead of the midpoint, which gives a line that is perpendicular but does not cut AB in half. Check by testing the midpoint in your final equation. This kind of question often has 4 or 5 marks, with separate marks for the midpoint, the gradient and the final equation.
Method: 1. Find the midpoint by averaging the x-coordinates and the y-coordinates. 2. Find the gradient of AB and then the perpendicular gradient. 3. Substitute the midpoint into y=mx+c to find c. 4. Check that the midpoint lies on your final line.
Once you have the equation of a perpendicular line, exams often ask where it crosses an axis. It crosses the x-axis where y=0, so substitute y=0 and solve for x. For y=-1/3x+5, putting y=0 gives 1/3x=5, so x=15 and the point is (15, 0). It crosses the y-axis at (0, c), which you can read directly.
To find where two lines meet, solve their equations together. For 2x+3y=12 and 3x-2y=5, multiply the first by 2 and the second by 3 to get 4x+6y=24 and 9x-6y=15. Adding gives 13x=39, so x=3, and then y=2. Check by putting (3, 2) into both original equations. The two lines are perpendicular, because their gradients -2/3 and 3/2 multiply to -1, so they cross at a right angle at (3, 2).
Method: 1. For an axis crossing, put y=0 or x=0 into the equation of the line. 2. For a meeting point, make one pair of coefficients match and add or subtract. 3. Find the second coordinate by substituting back. 4. Check the point in both original equations.
To show that triangle ABC has a right angle at B, find the gradient of AB and the gradient of BC, then multiply. For A(0, 1), B(4, 3) and C(2, 7) the gradients are (3-1)/(4-0)=1/2 and (7-3)/(2-4)=-2. Their product is 1/2×(-2)=-1, so the sides are perpendicular. Quote the product and state the conclusion in words: "so angle ABC is a right angle".
The same idea proves a quadrilateral is a rectangle, a kite or a trapezium. Parallel sides have equal gradients, and perpendicular sides have a product of -1. Once you know the angle is right, areas follow from lengths found with Pythagoras. Here AB=√(4^2+2^2)=√(20) and BC=√(2^2+4^2)=√(20), so the area is 1/2×√(20)×√(20)=10. Work with exact values to avoid rounding.
Method: 1. Work out the gradient of each side with subtraction in the same order. 2. Multiply the gradients of the two sides that meet at the angle. 3. State that a product of -1 means a right angle, naming the angle. 4. Use Pythagoras on the two sides to find lengths or areas.
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