Hard GCSE Parallel and perpendicular lines Questions

Challenging, exam-style GCSE Parallel and perpendicular lines questions with worked solutions. Stretch yourself on the hardest perpendicular bisector, midpoint, negative reciprocal, fractional gradient problems.

perpendicular bisectormidpointnegative reciprocalfractional gradientgradient between two pointsperpendicular lines
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
AA is the point (1,2)(1, 2) and BB is the point (5,10)(5, 10). The perpendicular bisector of ABAB crosses the xx-axis at PP. Work out the area of triangle APBAPB.
Show worked solution

Worked solution

  1. Find the midpoint of AB

    M=(1+52, 2+102)=(3,6)M = \left(\frac{1 + 5}{2},\ \frac{2 + 10}{2}\right) = (3, 6)

    The perpendicular bisector passes through the midpoint M of AB.

  2. Find the gradient of AB

    mAB=10−25−1=84=2m_{AB} = \frac{10 - 2}{5 - 1} = \frac{8}{4} = 2

    AB has gradient 2.

  3. Use the perpendicular rule

    m1×m2=−1m_1 \times m_2 = -1

    The bisector is perpendicular to AB.

  4. Find the gradient of the bisector

    m=−12m = -\frac{1}{2}

    The negative reciprocal of 2 is -1/2.

  5. Start the equation of the bisector

    y=−12x+cy = -\frac{1}{2}x + c

    Use y = mx + c with gradient -1/2.

  6. Substitute the midpoint

    6=−12×3+c=−32+c6 = -\frac{1}{2} \times 3 + c = -\frac{3}{2} + c

    The bisector passes through M(3, 6).

  7. Solve for c

    c=6+32=152c = 6 + \frac{3}{2} = \frac{15}{2}

    Add three halves to 6.

  8. Write the equation of the bisector

    y=−12x+152y = -\frac{1}{2}x + \frac{15}{2}

    This is the perpendicular bisector of AB.

  9. Find where it crosses the x-axis

    0=−12x+1520 = -\frac{1}{2}x + \frac{15}{2}

    On the x-axis, y=0y = 0.

  10. Solve for x

    x=15⇒P=(15,0)x = 15 \quad \Rightarrow \quad P = (15, 0)

    Multiply through by 2 to get 0=−x+150 = -x + 15.

  11. Find the length of AB

    AB=42+82=16+64=80AB = \sqrt{4^2 + 8^2} = \sqrt{16 + 64} = \sqrt{80}

    Use Pythagoras between A(1, 2) and B(5, 10). This is the base of the triangle.

  12. Find the height of the triangle

    PM=(15−3)2+(0−6)2=144+36=180PM = \sqrt{(15 - 3)^2 + (0 - 6)^2} = \sqrt{144 + 36} = \sqrt{180}

    PM is perpendicular to AB, so it is the height from P to the base AB.

  13. Use the area formula

    area=12×AB×PM\text{area} = \frac{1}{2} \times AB \times PM

    The base is AB and the height is the perpendicular distance PM.

  14. Substitute the lengths

    area=12×80×180=1214400\text{area} = \frac{1}{2} \times \sqrt{80} \times \sqrt{180} = \frac{1}{2}\sqrt{14400}

    80×180=1440080 \times 180 = 14400.

  15. Work out the area

    area=12×120=60\text{area} = \frac{1}{2} \times 120 = 60

    The square root of 14400 is 120, so the area of triangle APB is 60 square units.

Answer
6060
Question 2
5 markschallenging
The line 3x+ky=73x + ky = 7 is perpendicular to the line 2x−3y=12x - 3y = 1. Work out the value of kk.
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Worked solution

  1. Rearrange the first line

    ky=−3x+7ky = -3x + 7

    Subtract 3x from both sides.

  2. Divide by k

    y=−3kx+7ky = -\frac{3}{k}x + \frac{7}{k}

    Assume k is not zero, so that the line has a gradient.

  3. Write down the first gradient

    m1=−3km_1 = -\frac{3}{k}

    The coefficient of x is -3/k.

  4. Rearrange the second line

    −3y=−2x+1-3y = -2x + 1

    Subtract 2x from both sides of 2x−3y=12x - 3y = 1.

  5. Divide by −3-3

    y=23x−13y = \frac{2}{3}x - \frac{1}{3}

    Dividing by -3 changes every sign.

  6. Write down the second gradient

    m2=23m_2 = \frac{2}{3}

    The second line has gradient 2/3.

  7. Use the perpendicular rule

    m1×m2=−1m_1 \times m_2 = -1

    The two lines meet at a right angle.

  8. Substitute the gradients

    −3k×23=−1-\frac{3}{k} \times \frac{2}{3} = -1

    Put both gradients into the rule.

  9. Simplify the left-hand side

    −63k=−2k-\frac{6}{3k} = -\frac{2}{k}

    Cancel the factor of 3.

  10. Form the equation

    −2k=−1-\frac{2}{k} = -1

    The simplified product must equal -1.

  11. Multiply both sides by k

    −2=−k-2 = -k

    Clear the fraction.

  12. Solve for k

    k=2k = 2

    Multiply both sides by -1.

  13. Write the first line with k=2k = 2

    3x+2y=7⇒y=−32x+723x + 2y = 7 \Rightarrow y = -\frac{3}{2}x + \frac{7}{2}

    Substituting k=2k = 2 gives a gradient of -3/2.

  14. Check the perpendicular rule

    −32×23=−1-\frac{3}{2} \times \frac{2}{3} = -1

    The gradients multiply to -1, so the lines are perpendicular.

  15. State the answer

    k=2k = 2

    The value of k is 2.

Answer
k=2k = 2
Question 3
6 markschallenging
ABCDABCD has vertices A(0,1)A(0, 1), B(3,2)B(3, 2), C(4,5)C(4, 5) and D(1,4)D(1, 4). Show that the diagonals are perpendicular and use them to work out the area of ABCDABCD.
Show worked solution

Worked solution

  1. Write down the distance formula

    d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

    First check the four sides are the same length.

  2. Find AB and BC

    AB=32+12=10,BC=12+32=10AB = \sqrt{3^2 + 1^2} = \sqrt{10}, \quad BC = \sqrt{1^2 + 3^2} = \sqrt{10}

    Both sides have length root 10.

  3. Find CD and DA

    CD=(−3)2+(−1)2=10,DA=(−1)2+(−3)2=10CD = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10}, \quad DA = \sqrt{(-1)^2 + (-3)^2} = \sqrt{10}

    All four sides are equal, so ABCD is a rhombus.

  4. Write down the gradient formula

    m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}

    Now look at the diagonals AC and BD.

  5. Find the gradient of AC

    mAC=5−14−0=44=1m_{AC} = \frac{5 - 1}{4 - 0} = \frac{4}{4} = 1

    Diagonal AC has gradient 1.

  6. Find the gradient of BD

    mBD=4−21−3=2−2=−1m_{BD} = \frac{4 - 2}{1 - 3} = \frac{2}{-2} = -1

    Diagonal BD has gradient -1.

  7. Multiply the gradients

    1×(−1)=−11 \times (-1) = -1

    The product is -1, so the diagonals are perpendicular.

  8. Find the length of AC

    AC=42+42=32=42AC = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}

    Use Pythagoras between A(0, 1) and C(4, 5).

  9. Find the length of BD

    BD=(−2)2+22=8=22BD = \sqrt{(-2)^2 + 2^2} = \sqrt{8} = 2\sqrt{2}

    Use Pythagoras between B(3, 2) and D(1, 4).

  10. Recall the area rule for perpendicular diagonals

    area=12×d1×d2\text{area} = \frac{1}{2} \times d_1 \times d_2

    This works for any quadrilateral whose diagonals cross at right angles.

  11. Substitute the diagonals

    area=12×42×22\text{area} = \frac{1}{2} \times 4\sqrt{2} \times 2\sqrt{2}

    Put the two lengths into the formula.

  12. Multiply the surds

    42×22=8×2=164\sqrt{2} \times 2\sqrt{2} = 8 \times 2 = 16

    Root 2 times root 2 is 2.

  13. Work out the area

    area=12×16=8\text{area} = \frac{1}{2} \times 16 = 8

    The area of ABCD is 8 square units.

  14. Check the diagonals bisect each other

    midpoint AC=(2,3),midpoint BD=(2,3)\text{midpoint } AC = (2, 3), \quad \text{midpoint } BD = (2, 3)

    Both diagonals have the same midpoint, so they bisect each other, as in any rhombus.

  15. State the area

    area=8\text{area} = 8

    ABCD is a rhombus with perpendicular diagonals and an area of 8 square units.

Answer
88
Question 4
6 markschallenging
Work out the shortest distance from the origin to the line 3x+4y=123x + 4y = 12.
Show worked solution

Worked solution

  1. Rearrange the line

    4y=−3x+124y = -3x + 12

    Subtract 3x from both sides.

  2. Divide by 4

    y=−34x+3y = -\frac{3}{4}x + 3

    The line is now in the form y = mx + c.

  3. Read off the gradient

    m1=−34m_1 = -\frac{3}{4}

    The line has gradient -3/4.

  4. Find the perpendicular gradient

    m2=43m_2 = \frac{4}{3}

    The negative reciprocal of -3/4 is 4/3.

  5. Write the perpendicular through the origin

    y=43xy = \frac{4}{3}x

    The shortest distance is measured along the perpendicular from the origin, so c=0c = 0.

  6. Substitute into the line

    3x+4×43x=123x + 4 \times \frac{4}{3}x = 12

    Put y=4x/3y = 4x/3 into 3x+4y=123x + 4y = 12 to find where the two lines cross.

  7. Multiply through by 3

    9x+16x=369x + 16x = 36

    Clearing the fraction gives whole numbers.

  8. Collect the x terms

    25x=3625x = 36

    Add the like terms on the left.

  9. Solve for x

    x=3625x = \frac{36}{25}

    Divide both sides by 25.

  10. Find y

    y=43×3625=4825y = \frac{4}{3} \times \frac{36}{25} = \frac{48}{25}

    The foot of the perpendicular is (36/25, 48/25).

  11. Write down the distance formula

    d=x2+y2d = \sqrt{x^2 + y^2}

    Use Pythagoras from the origin to the foot of the perpendicular.

  12. Substitute the coordinates

    d=(3625)2+(4825)2d = \sqrt{\left(\frac{36}{25}\right)^2 + \left(\frac{48}{25}\right)^2}

    Square each coordinate.

  13. Add the squares

    d=1296+2304625=3600625d = \sqrt{\frac{1296 + 2304}{625}} = \sqrt{\frac{3600}{625}}

    36 squared is 1296 and 48 squared is 2304.

  14. Take the square root

    d=6025=125d = \frac{60}{25} = \frac{12}{5}

    The square root of 3600 is 60 and the square root of 625 is 25.

  15. State the answer

    d=2.4d = 2.4

    The shortest distance from the origin to the line is 2.4 units.

Answer
2.42.4
Question 5
5 markschallenging
Three sides of a rectangle lie on the lines y=2x+1y = 2x + 1, y=2x−4y = 2x - 4 and y=−12x+3y = -\frac{1}{2}x + 3. The fourth side passes through the point (4,−3)(4, -3). Find the equation of the line containing the fourth side.
Show worked solution

Worked solution

  1. Compare the first two lines

    m=2 and m=2m = 2 \text{ and } m = 2

    They have equal gradients, so they are parallel: these are two opposite sides.

  2. Look at the third line

    m=−12m = -\frac{1}{2}

    Its gradient is different, so it is one of the other pair of sides.

  3. Check the third line is perpendicular to the first pair

    2×(−12)=−12 \times \left(-\frac{1}{2}\right) = -1

    The product of the gradients is -1, so the corners are right angles.

  4. Decide the gradient of the fourth side

    m4=−12m_4 = -\frac{1}{2}

    The fourth side is opposite the third, so it is parallel to it.

  5. Check the fourth side is perpendicular to the parallel pair

    2×(−12)=−12 \times \left(-\frac{1}{2}\right) = -1

    It must also meet y=2x+1y = 2x + 1 and y=2x−4y = 2x - 4 at right angles, and it does.

  6. Start the equation of the fourth side

    y=−12x+cy = -\frac{1}{2}x + c

    Use y = mx + c with gradient -1/2.

  7. Substitute the given point

    −3=−12×4+c-3 = -\frac{1}{2} \times 4 + c

    The fourth side passes through (4, -3).

  8. Work out the product

    −12×4=−2-\frac{1}{2} \times 4 = -2

    Half of 4 is 2.

  9. Simplify the equation

    −3=−2+c-3 = -2 + c

    Substitute the product back in.

  10. Solve for c

    c=−3+2=−1c = -3 + 2 = -1

    Add 2 to both sides.

  11. Write the equation

    y=−12x−1y = -\frac{1}{2}x - 1

    This is the fourth side of the rectangle.

  12. Check the point

    y=−12×4−1=−3y = -\frac{1}{2} \times 4 - 1 = -3

    Substituting x=4x = 4 gives y=−3y = -3, so (4, -3) lies on the line.

  13. Check it is different from the third side

    −1≠3-1 \neq 3

    The intercepts differ, so it is a genuinely different line, parallel to the third side.

  14. Check the shape closes up

    two pairs of parallel sides, all angles 90∘\text{two pairs of parallel sides, all angles } 90^\circ

    The four lines form a rectangle.

  15. State the answer

    y=−12x−1y = -\frac{1}{2}x - 1

    The fourth side lies on this line.

Answer
y=−12x−1y = -\frac{1}{2}x - 1

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