A-Level Straight lines Practice Questions

Free A-Level Straight lines practice questions with full step-by-step worked solutions. Covers gradient, coordinate geometry, midpoint, distance. Practise exam-style problems and check your method.

gradientcoordinate geometrymidpointdistancePythagorasequation of a line
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
Find the gradient of the line joining the points A(1, 2)A\left(1,\ 2\right) and B(4, 11)B\left(4,\ 11\right).
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Worked solution

  1. Write down the two points

    A(1, 2),B(4, 11)A\left(1,\ 2\right),\quad B\left(4,\ 11\right)

    Label the first point (x_1, y_1) and the second (x_2, y_2). Keeping them in order stops sign mistakes later.

  2. Recall the gradient formula

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

    The gradient measures steepness: how much the line goes up (or down) for each step to the right. This is the change in y divided by the change in x.

  3. Substitute the coordinates

    m=11−(2)4−(1)m=\frac{11-(2)}{4-(1)}

    Carefully put each number in its place. Subtracting a negative becomes adding, so watch those double signs.

  4. Simplify to find the gradient

    m=3m=3

    Work out the top and bottom, then simplify the fraction. A positive answer slopes uphill, a negative one slopes downhill.

Answer
33
Question 2
3 markseasy
Find the equation of the line passing through A(1, 1)A\left(1,\ 1\right) and B(3, 7)B\left(3,\ 7\right), giving your answer in the form y=mx+cy = mx + c.
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Worked solution

  1. Find the gradient

    m=7−(1)3−(1)=3m=\frac{7-(1)}{3-(1)}=3

    With two points we first work out the gradient using change in y over change in x. This tells us the steepness of the line through them.

  2. Substitute into y - y_1 = m(x - x_1)

    y−(1)=3(x−(1))y-(1)=3(x-(1))

    Use the gradient and one of the points. It does not matter which point you pick; both give the same line.

  3. Expand and simplify

    y=3x−2y=3x-2

    Multiply out and tidy up to reach y = mx + c. This is the equation of the straight line through both points.

Answer
y=3x−2y=3x-2
Question 3
4 marksintermediate
Find the equation of the perpendicular bisector of the line segment joining A(1, 4)A\left(1,\ 4\right) and B(7, 2)B\left(7,\ 2\right), giving your answer in the form ax+by+c=0ax+by+c=0.
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Worked solution

  1. Find the midpoint of AB

    M=(1+72, 4+22)=(4, 3)M=\left(\frac{1+7}{2},\ \frac{4+2}{2}\right)=\left(4,\ 3\right)

    The perpendicular bisector cuts AB exactly in half, so it must pass through the midpoint. Average the coordinates to find it.

  2. Find the gradient of AB

    mAB=2−(4)7−(1)=−13m_{AB}=\frac{2-(4)}{7-(1)}=-\frac{1}{3}

    We need the gradient of AB so we can turn it perpendicular. Change in y over change in x as usual.

  3. Find the perpendicular gradient

    m′=−1−13=3m'=-\frac{1}{-\frac{1}{3}}=3

    The bisector is at right angles to AB, so we take the negative reciprocal of the gradient of AB.

  4. Substitute M and m' into y - y_1 = m(x - x_1)

    y−(3)=3(x−(4))y-(3)=3(x-(4))

    Use the midpoint and the perpendicular gradient together. This fixes the bisector in place.

  5. Expand the bracket

    y−3=3x−12y-3=3x-12

    Multiply out and be careful with fractions and signs. Take it one term at a time.

  6. Give the final equation

    3x−y−9=03x-y-9=0

    Rearranging gives the equation of the perpendicular bisector of AB. Every point on it is equally far from A and B.

Answer
3x−y−9=03x-y-9=0
Question 4
6 markshard
A triangle has vertices A(−1, 6)A\left(-1,\ 6\right), B(2, 0)B\left(2,\ 0\right) and C(6, 4)C\left(6,\ 4\right). Find the equation of the median from AA, in the form ax+by+c=0ax+by+c=0.
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Worked solution

  1. Write down the three vertices

    A(−1, 6), B(2, 0), C(6, 4)A\left(-1,\ 6\right),\ B\left(2,\ 0\right),\ C\left(6,\ 4\right)

    List the vertices so we know exactly which point the median starts from and which side is opposite.

  2. Recall what a median is

    median from A→midpoint of BC\text{median from }A\to\text{midpoint of }BC

    A median joins a vertex to the midpoint of the opposite side. Here it goes from A to the midpoint of BC.

  3. Identify the opposite side

    side opposite A is BC\text{side opposite }A\text{ is }BC

    The median from A must reach the far side, which is BC. So the first job is to find the midpoint of BC.

  4. Find the midpoint of BC

    M=(2+62, 0+42)=(4, 2)M=\left(\frac{2+6}{2},\ \frac{0+4}{2}\right)=\left(4,\ 2\right)

    Average B and C to find the midpoint of the opposite side.

  5. Find the gradient of AM

    m=2−(6)4−(−1)=−45m=\frac{2-(6)}{4-(-1)}=-\frac{4}{5}

    The median passes through A and M, so use these two points for the gradient.

  6. Substitute into y - y_1 = m(x - x_1)

    y−(6)=−45(x−(−1))y-(6)=-\frac{4}{5}(x-(-1))

    Use vertex A and the gradient just found to build the equation.

  7. Multiply through to clear the fraction

    5(y−(6))=−4(x−(−1))5\big(y-(6)\big)=-4\big(x-(-1)\big)

    Multiplying both sides by the denominator of the gradient removes the fraction, ready for integer coefficients.

  8. Expand both sides

    −4x−5y+26=0-4x-5y+26=0

    Multiply out the brackets and gather every term on one side so the other side is 0.

  9. Tidy to general form ax + by + c = 0

    4x+5y−26=04x+5y-26=0

    Divide by any common factor and make the x-coefficient positive. This is the tidy general form.

  10. Check both A and M satisfy the equation

    A(−1, 6) and M(4, 2) both satisfy 4x+5y−26=0A\left(-1,\ 6\right)\ \text{and}\ M\left(4,\ 2\right)\ \text{both satisfy }4x+5y-26=0

    Substituting A and M should each give 0. They do, confirming the median passes through both points.

Answer
4x+5y−26=04x+5y-26=0
Question 5
8 markschallenging
Find the orthocentre (the intersection of the altitudes) of the triangle with vertices A(1, 1)A\left(1,\ 1\right), B(5, 1)B\left(5,\ 1\right) and C(2, 5)C\left(2,\ 5\right).
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Worked solution

  1. Understand the goal

    orthocentre = intersection of the altitudes\text{orthocentre = intersection of the altitudes}

    An altitude goes from a vertex perpendicular to the opposite side. The three altitudes meet at the orthocentre; we use two of them.

  2. Find the gradient of the line joining the two points

    mBC=y2−y1x2−x1=5−(1)2−(5)=−43m_{BC}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-(1)}{2-(5)}=-\frac{4}{3}

    The gradient tells us how steep the line is. We subtract the y-values on top and the x-values underneath, always keeping the points in the same order. Remember from earlier work: rise over run.

  3. Find the perpendicular gradient

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

    Perpendicular lines have gradients that multiply to give -1, so we flip the fraction and change the sign. This is the negative reciprocal rule.

  4. Substitute the gradient and point into y - y_1 = m(x - x_1)

    y−(1)=34(x−(1))y-(1)=\frac{3}{4}(x-(1))

    We use the point-gradient form because we know one point the line passes through and its gradient. Putting the numbers in fixes the line in place.

  5. Rearrange into the form y = mx + c

    y=34x+14y=\frac{3}{4}x+\frac{1}{4}

    Expanding the bracket and tidying up gives the familiar straight-line equation. The number c is where the line crosses the y-axis.

  6. This is the altitude from A

    y=34x+14y=\frac{3}{4}x+\frac{1}{4}

    It passes through A and is perpendicular to BC, so it is the altitude from A.

  7. Find the gradient of the line joining the two points

    mAC=y2−y1x2−x1=5−(1)2−(1)=4m_{AC}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-(1)}{2-(1)}=4

    The gradient tells us how steep the line is. We subtract the y-values on top and the x-values underneath, always keeping the points in the same order. Remember from earlier work: rise over run.

  8. Find the perpendicular gradient

    mB=−14=−14m_B=-\frac{1}{4}=-\frac{1}{4}

    Perpendicular lines have gradients that multiply to give -1, so we flip the fraction and change the sign. This is the negative reciprocal rule.

  9. Substitute the gradient and point into y - y_1 = m(x - x_1)

    y−(1)=−14(x−(5))y-(1)=-\frac{1}{4}(x-(5))

    We use the point-gradient form because we know one point the line passes through and its gradient. Putting the numbers in fixes the line in place.

  10. Rearrange into the form y = mx + c

    y=−14x+94y=-\frac{1}{4}x+\frac{9}{4}

    Expanding the bracket and tidying up gives the familiar straight-line equation. The number c is where the line crosses the y-axis.

  11. This is the altitude from B

    y=−14x+94y=-\frac{1}{4}x+\frac{9}{4}

    It passes through B and is perpendicular to AC, so it is the altitude from B.

  12. Solve the two altitudes together

    34x+14=−14x+94\frac{3}{4}x+\frac{1}{4}=-\frac{1}{4}x+\frac{9}{4}

    Where two lines cross, they share the same x and y. Setting the two right-hand sides equal lets us find that shared x-value first.

  13. Solve for x

    x=2x=2

    Collect the x-terms on one side and the numbers on the other, then divide to get x. Take your time with the signs here.

  14. Substitute back to find y

    y=74y=\frac{7}{4}

    Now put the x-value back into either equation to get y. Using the simpler equation reduces the chance of a slip.

  15. State the orthocentre

    (2, 74)\left(2,\ \frac{7}{4}\right)

    This point where the altitudes cross is the orthocentre of the triangle.

Answer
(2, 74)\left(2,\ \frac{7}{4}\right)

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