GCSE Parallel and perpendicular lines Practice Questions

Free GCSE Parallel and perpendicular lines practice questions with full step-by-step worked solutions. Covers parallel lines, equal gradients, y = mx + c, identifying gradients. Practise exam-style problems and check your method.

parallel linesequal gradientsy = mx + cidentifying gradientsnegative gradientperpendicular lines
GCSE Foundation70 questionsStep-by-step solutions
Question 1
1 markeasy
A straight line is parallel to the line with equation y=3x+4y = 3x + 4. Write down the gradient of the parallel line.
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Worked solution

  1. Compare with y=mx+cy = mx + c

    y=3x+4y = 3x + 4

    In the form y = mx + c the number multiplying x is the gradient.

  2. Write down the gradient of the given line

    m=3m = 3

    The coefficient of x is 3, so the given line has gradient 3.

  3. Use the rule for parallel lines

    mparallel=3m_{\text{parallel}} = 3

    Parallel lines never meet, so they have exactly the same gradient.

Answer
33
Question 2
2 markseasy
A line passes through the origin and is perpendicular to y=6x−5y = 6x - 5. Write down its equation.
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Worked solution

  1. Find the gradient of the given line

    y=6x−5⇒m1=6y = 6x - 5 \Rightarrow m_1 = 6

    The gradient of the given line is 6.

  2. Find the perpendicular gradient

    m2=−16=−16m_2 = \frac{-1}{6} = -\frac{1}{6}

    The negative reciprocal of 6 is -1/6.

  3. Use the origin as the intercept

    c=0⇒y=−16xc = 0 \Rightarrow y = -\frac{1}{6}x

    A line through the origin has y-intercept 0.

Answer
y=−16xy = -\frac{1}{6}x
Question 3
2 marksintermediate
Find the equation of the line parallel to y=−4x+1y = -4x + 1 that passes through the point (−2,3)(-2, 3).
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Worked solution

  1. Find the gradient of the given line

    y=−4x+1y = -4x + 1

    Comparing with y = mx + c, the gradient is -4.

  2. Use equal gradients for parallel lines

    m=−4m = -4

    Parallel lines have the same gradient, so the new line also has this gradient.

  3. Start the equation of the new line

    y=−4x+cy = -4x + c

    Only the intercept c is still unknown.

  4. Substitute the given point

    3=−4×−2+c=8+c3 = -4 \times -2 + c = 8 + c

    The line passes through (-2, 3), so x=−2x = -2 and y=3y = 3 must fit the equation.

  5. Solve for c

    c=3−8=−5c = 3 - 8 = -5

    Rearranging gives the y-intercept of the new line.

  6. Write the equation

    y=−4x−5y = -4x - 5

    This line has the same gradient as the given line and passes through the given point.

Answer
y=−4x−5y = -4x - 5
Question 4
4 markshard
The line 3x+4y=243x + 4y = 24 crosses the xx-axis at AA and the yy-axis at BB. Work out the gradient of a line perpendicular to ABAB.
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Worked solution

  1. Find where the line crosses the x-axis

    3x+4×0=243x + 4 \times 0 = 24

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

  2. Solve for x

    x=8⇒A=(8,0)x = 8 \quad \Rightarrow \quad A = (8, 0)

    Divide both sides by 3.

  3. Find where the line crosses the y-axis

    3×0+4y=243 \times 0 + 4y = 24

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

  4. Solve for y

    y=6⇒B=(0,6)y = 6 \quad \Rightarrow \quad B = (0, 6)

    Divide both sides by 4.

  5. Write down the gradient formula

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

    Use the points A(8, 0) and B(0, 6).

  6. Substitute the coordinates

    mAB=6−00−8m_{AB} = \frac{6 - 0}{0 - 8}

    Change in y is 6; change in x is -8.

  7. Simplify the gradient

    mAB=−68=−34m_{AB} = -\frac{6}{8} = -\frac{3}{4}

    AB has gradient -3/4, which agrees with rearranging to y=−3/4y = -3/4 x + 6.

  8. Use the perpendicular rule

    m1×m2=−1m_1 \times m_2 = -1

    Perpendicular gradients multiply to -1.

  9. Take the negative reciprocal

    m=−1−34=43m = \frac{-1}{-\frac{3}{4}} = \frac{4}{3}

    Turn -3/4 upside down to get -4/3, then change the sign.

  10. Check the rule

    −34×43=−1-\frac{3}{4} \times \frac{4}{3} = -1

    The product is -1, so the gradient 4/3 is correct.

Answer
43\frac{4}{3}
Question 5
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.
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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

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