Parallel and Perpendicular Lines

Parallel non-vertical lines have equal gradients.

1. In y=mx+c the gradient is m, the number multiplying x. So y=4x-7 has gradient 4.
2. Parallel lines have the same gradient, so any line parallel to it has gradient 4.

The product of two non-vertical perpendicular gradients is minus one.

1. The line y=2x+9 has gradient 2.
2. Perpendicular gradients multiply to -1, so the new gradient is the negative reciprocal of 2: turn 2 upside down to get \frac12, then change the sign.
3. Check: 2\times\left(-\frac12\right)=-1. The gradient is -\frac12.

Parallel lines have the same gradient: compare the number in front of x.

1. Parallel lines have the same gradient. The line y=-3x+1 has gradient -3.
2. Only y=-3x-5 also has gradient -3, so it is the parallel line.
3. y=3x+1 has gradient 3: the sign matters. y=\frac13x-5 is perpendicular to the given line, not parallel.

Keep the gradient and use the new y-intercept.

1. The parallel line has gradient 5.
2. Its point (0,6) gives y-intercept 6, so its equation is y=5x+6.

A line perpendicular to a horizontal line is vertical.

1. The line y=-3 is horizontal.
2. A line perpendicular to a horizontal line is vertical, and every point on a vertical line has the same x-coordinate.
3. This line passes through (4,1), so its equation is x=4.

Make y the subject of 3x-2y=7 to find its gradient.

1. Make y the subject: 2y=3x-7, so y=\frac32x-\frac72. The gradient is \frac32.
2. A parallel line has the same gradient, so its equation is y=\frac32x+c.
3. Substitute the point (2,5): 5=\frac32\times2+c, so 5=3+c and c=2.
4. The equation is y=\frac32x+2.

Use the negative reciprocal gradient, then substitute the point.

1. The given line has gradient -\frac23.
2. A perpendicular line has the negative reciprocal gradient: \frac32. Check: -\frac23\times\frac32=-1.
3. Substitute (4,1) into y=\frac32x+c: 1=6+c, so c=-5.
4. The line is y=\frac32x-5.

Multiply each pair and look for a product of minus one.

1. Two lines are perpendicular when their gradients multiply to -1.
2. (-2)\times\frac12=-1, so -2 and \frac12 are perpendicular gradients.
3. -2 and -\frac12 are reciprocals but have the same sign, so their product is +1: the sign must change too. The other products are -4 and -\frac14.

Make y the subject of the second equation to find its gradient.

1. Divide 3y=12x+7 by 3: y=4x+\frac73, so its gradient is 4.
2. The first line has gradient k-2. Parallel lines have equal gradients, so k-2=4.
3. Therefore k=6.

Rearrange the first equation, then use the perpendicular-gradient product.

1. Make y the subject of ax+4y=8: 4y=-ax+8, so y=-\frac a4x+2. Its gradient is -\frac a4.
2. The line y=-2x+5 has gradient -2, so a perpendicular line has gradient \frac12 (the negative reciprocal).
3. So -\frac a4=\frac12. Multiply both sides by -4: a=-2.

Work out the gradient of AB; a parallel line has the same gradient.

1. Gradient =\frac{\text{change in }y}{\text{change in }x}, so AB has gradient \frac{6-3}{4-(-2)}=\frac36=\frac12.
2. L is parallel to AB, so it also has gradient \frac12: y=\frac12x+c.
3. Substitute (2,-1): -1=1+c, so c=-2.
4. L has equation y=\frac12x-2.

Use the two given points to find the original gradient before taking its negative reciprocal.

1. Gradient of the line through (-1,2) and (5,-1): \frac{-1-2}{5-(-1)}=\frac{-3}{6}=-\frac12.
2. The perpendicular gradient is the negative reciprocal of -\frac12, which is 2.
3. Substitute (3,1) into y=2x+c: 1=6+c, so c=-5.
4. The equation is y=2x-5.

Make y the subject of 2x+5y=20 to find the gradient of L.

1. Make y the subject: 5y=-2x+20, so y=-\frac25x+4. The gradient of L is -\frac25.
2. The perpendicular gradient is the negative reciprocal: \frac52. Check: -\frac25\times\frac52=-1.
3. Substitute (4,-3) into y=\frac52x+c: -3=10+c, so c=-13.
4. The equation is y=\frac52x-13 (or 5x-2y=26).

Work out both gradients, taking the points in the same order each time.

1. Gradient of AB: \frac{3-1}{5-1}=\frac24=\frac12.
2. Gradient of BC: \frac{5-3}{4-5}=\frac{2}{-1}=-2.
3. \frac12\times(-2)=-1, and gradients that multiply to -1 belong to perpendicular lines. So the last statement is the right one.
4. Gradients that add to 0, or are equal, do not show lines are perpendicular: equal gradients mean the lines are parallel.

A perpendicular bisector passes through the midpoint and has a perpendicular gradient.

1. The perpendicular bisector of AB passes through the midpoint of AB and is perpendicular to AB.
2. Midpoint: \left(\frac{-3+5}{2},\frac{2+6}{2}\right)=(1,4).
3. Gradient of AB: \frac{6-2}{5-(-3)}=\frac48=\frac12, so the perpendicular gradient is -2.
4. Substitute (1,4) into y=-2x+c: 4=-2+c, so c=6.
5. The perpendicular bisector is y=-2x+6.

Find each line equation before solving them together.

Part (a)

1. Parallel lines have equal gradients, so L has gradient 2: y=2x+c.
2. Substitute (2,7): 7=4+c, so c=3.
3. L has equation y=2x+3.

Part (b)

1. M is perpendicular to L, so its gradient is the negative reciprocal of 2, which is -\frac12.
2. (0,13) is on the y-axis, so the y-intercept is 13. M has equation y=-\frac12x+13.

Part (c)

1. Where the lines meet, both equations give the same y: 2x+3=-\frac12x+13.
2. Add \frac12x and subtract 3: \frac52x=10, so x=4.
3. Then y=2\times4+3=11. The lines intersect at (4,11).

Find the equation of the line through P that is perpendicular to L, then solve it together with the equation of L.

Part (a)

1. L has gradient 3, so PF has gradient -\frac13 (the negative reciprocal).
2. Line PF: y=-\frac13x+c. Substitute P(7,-1): -1=-\frac73+c, so c=\frac43.
3. F is on both lines, so 3x-2=-\frac13x+\frac43. Multiply by 3: 9x-6=-x+4.
4. So 10x=10, giving x=1, and y=3\times1-2=1. F is (1,1).

Part (b)

1. From F(1,1) to P(7,-1) is 6 across and 2 down.
2. By Pythagoras, PF=\sqrt{6^2+2^2}=\sqrt{40}.
3. \sqrt{40}=6.324\ldots, so PF=6.32 to 3 significant figures.

Write the gradient of CD in terms of k and set it equal to the gradient of AB.

1. Gradient of AB: \frac{5-1}{4-(-2)}=\frac46=\frac23.
2. Gradient of CD: \frac{1-k}{k-0}=\frac{1-k}{k}.
3. Parallel lines have equal gradients: \frac{1-k}{k}=\frac23. Multiply both sides by 3k: 3(1-k)=2k.
4. So 3-3k=2k, which gives 3=5k and k=\frac35 (or 0.6).

Find the gradient and y-intercept of the parallel line before constructing the perpendicular line.

1. Make y the subject of 2x+3y=9: y=-\frac23x+3. So L, which is parallel to it, has gradient -\frac23.
2. L passes through (6,0): 0=-\frac23\times6+c, so 0=-4+c and c=4. L is y=-\frac23x+4.
3. So L crosses the y-axis at (0,4).
4. M is perpendicular to L, so its gradient is the negative reciprocal of -\frac23, which is \frac32.
5. M passes through (0,4), so its y-intercept is 4: M is y=\frac32x+4.

Adjacent sides of a rectangle are perpendicular; opposite sides have the same coordinate changes.

Part (a)

1. Gradient of AB: \frac{3-1}{4-0}=\frac12.
2. The corners of a rectangle are right angles, so BC is perpendicular to AB and has gradient -2.
3. Line BC: y=-2x+c through B(4,3): 3=-8+c, so c=11 and BC is y=-2x+11.
4. At C, y=8: 8=-2x+11, so x=\frac32. C is \left(\frac32,8\right).

Part (b)

1. Opposite sides of a rectangle are equal and parallel, so the move from D to C is the same as the move from A to B: 4 right and 2 up.
2. So D is 4 left and 2 down from C: D=\left(\frac32-4,\ 8-2\right)=\left(-\frac52,6\right).