Hard A-Level Straight lines Questions

Challenging, exam-style A-Level Straight lines questions with worked solutions. Stretch yourself on the hardest foot of perpendicular, intersection, perpendicular lines, shortest distance problems.

foot of perpendicularintersectionperpendicular linesshortest distanceperpendicularparallelogram
A-Level34 questionsStep-by-step solutions
Question 1
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)
Question 2
8 markschallenging
A point XX moves along the line y=3x−4y=3x-4. Find the exact minimum possible distance from XX to the point P(−2, 5)P\left(-2,\ 5\right).
Show worked solution

Worked solution

  1. Interpret 'minimum distance'

    min⁡PX occurs when PX⊥ℓ\min PX\ \text{occurs when }PX\perp\ell

    The closest point X on a line to P is the foot of the perpendicular from P. Any other point on the line is further away.

  2. Write down the point and line

    P(−2, 5),y=3x−4P\left(-2,\ 5\right),\quad y=3x-4

    Note the fixed point P and the line X moves along. This keeps the numbers organised.

  3. Find the gradient of the line

    m=3m=3

    Read the gradient of the given line.

  4. Find the perpendicular gradient

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

    The shortest path is perpendicular to the line, so take the negative reciprocal.

  5. Perpendicular through P (point-gradient form)

    y−(5)=−13(x−(−2))y-(5)=-\frac{1}{3}(x-(-2))

    Use P and the perpendicular gradient to start the equation of the shortest-path line.

  6. Tidy the perpendicular

    y=−13x+133y=-\frac{1}{3}x+\frac{13}{3}

    Rearrange into y = mx + c so it is ready to solve with the given line.

  7. Set the two lines equal

    3x−4=−13x+1333x-4=-\frac{1}{3}x+\frac{13}{3}

    Solve simultaneously to find X, the foot of the perpendicular.

  8. Collect the x-terms

    103x=253\frac{10}{3}x=\frac{25}{3}

    Gather x-terms and constants on opposite sides, ready to divide.

  9. Solve for x of X

    x=52x=\frac{5}{2}

    Divide to find the foot's x-coordinate.

  10. Find y of X

    y=72 ⇒ X(52, 72)y=\frac{7}{2}\ \Rightarrow\ X\left(\frac{5}{2},\ \frac{7}{2}\right)

    Substitute back to find y. X is the closest point on the line to P.

  11. Check X lies on the line

    3(52)−4=723(\frac{5}{2})-4=\frac{7}{2}

    Substituting X into the line should hold true, confirming X is genuinely on it.

  12. Find the horizontal and vertical gaps

    Δx=92, Δy=−32\Delta x=\frac{9}{2},\ \Delta y=-\frac{3}{2}

    Measure how far X is from P across and up, ready for Pythagoras.

  13. Square and add

    PX2=(92)2+(−32)2=45PX^2=(\frac{9}{2})^2+(-\frac{3}{2})^2=45

    Square each gap and add to get the squared distance.

  14. Take the square root for the minimum distance

    PX=452PX=\frac{45}{2}

    The square root gives the exact minimum distance from P to the line.

  15. Confirm this is the minimum

    any other point on ℓ is further from P\text{any other point on }\ell\text{ is further from }P

    Because X is the foot of the perpendicular, every other point on the line forms a longer slanted distance to P. So this value really is the minimum.

Answer
452\frac{45}{2}
Question 3
8 markschallenging
Find the reflection of the point P(3, 1)P\left(3,\ 1\right) in the line y=13x+2y=\frac{1}{3}x+2.
Show worked solution

Worked solution

  1. Plan the reflection

    reflect P in the line y=13x+2\text{reflect }P\text{ in the line }y=\frac{1}{3}x+2

    To reflect a point in a line, drop a perpendicular from the point to the line, find the foot, then go the same distance again beyond it.

  2. Write down the point and mirror line

    P(3, 1),y=13x+2P\left(3,\ 1\right),\quad y=\frac{1}{3}x+2

    Note the point to be reflected and the mirror line. Keeping them clear avoids mixing up numbers later.

  3. Find the gradient of the mirror line

    m=13m=\frac{1}{3}

    Read the gradient of the line we are reflecting in.

  4. Find the perpendicular gradient

    m′=−113=−3m'=-\frac{1}{\frac{1}{3}}=-3

    The path from P to the line is perpendicular to it, so use the negative reciprocal.

  5. Equation of the perpendicular through P

    y−(1)=−3(x−(3))y-(1)=-3(x-(3))

    Use P and the perpendicular gradient in point-gradient form to get this line.

  6. Tidy the perpendicular

    y=−3x+10y=-3x+10

    Rearrange into y = mx + c so it is ready to solve with the mirror line.

  7. Set equal to the mirror line

    13x+2=−3x+10\frac{1}{3}x+2=-3x+10

    The foot of the perpendicular is where these two lines meet, so solve simultaneously.

  8. Collect the x-terms

    103x=8\frac{10}{3}x=8

    Gather x-terms and constants on opposite sides, ready to divide.

  9. Solve for x of the foot

    x=125x=\frac{12}{5}

    Divide to find the foot's x-coordinate.

  10. Find y of the foot

    y=145 ⇒ F(125, 145)y=\frac{14}{5}\ \Rightarrow\ F\left(\frac{12}{5},\ \frac{14}{5}\right)

    Substitute back to get the foot F, which is the midpoint of P and its image.

  11. Check F lies on the mirror line

    13(125)+2=145\frac{1}{3}(\frac{12}{5})+2=\frac{14}{5}

    Substituting F into the mirror line should hold true. It does, so F is genuinely on the line.

  12. Use F as the midpoint of P and its image (x)

    3+x′2=125 ⇒ x′=95\frac{3+x'}{2}=\frac{12}{5}\ \Rightarrow\ x'=\frac{9}{5}

    Since F is halfway between P and its reflection, double F's x and subtract P's x.

  13. Use F as the midpoint (y)

    1+y′2=145 ⇒ y′=235\frac{1+y'}{2}=\frac{14}{5}\ \Rightarrow\ y'=\frac{23}{5}

    Do the same for the y-coordinate to find the image's height.

  14. Check PF equals FP'

    PF2=185=FP′2PF^2=\frac{18}{5}=FP'^2

    P and its image should be the same distance from the foot F. The equal squared distances confirm this.

  15. State the reflected point

    (95, 235)\left(\frac{9}{5},\ \frac{23}{5}\right)

    This image point P' is the mirror image of P in the line. F is exactly midway between them.

Answer
(95, 235)\left(\frac{9}{5},\ \frac{23}{5}\right)
Question 4
8 markschallenging
Find the circumcentre of the triangle with vertices A(2, 1)A\left(2,\ 1\right), B(6, 3)B\left(6,\ 3\right) and C(3, 8)C\left(3,\ 8\right) (the point equidistant from all three).
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Worked solution

  1. Understand the goal

    circumcentre = centre of circle through A,B,C\text{circumcentre = centre of circle through }A,B,C

    The circumcentre is equidistant from all three vertices, so it lies on the perpendicular bisector of every side. We find two bisectors and cross them.

  2. Find the midpoint

    M=(2+62, 1+32)=(4, 2)M=\left(\frac{2+6}{2},\ \frac{1+3}{2}\right)=\left(4,\ 2\right)

    The midpoint sits exactly halfway between the two points, so we average the x-values and average the y-values. Adding and dividing by 2 finds the middle.

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

    mAB=y2−y1x2−x1=3−(1)6−(2)=12m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{3-(1)}{6-(2)}=\frac{1}{2}

    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.

  4. Find the perpendicular gradient

    m′=−112=−2m'=-\frac{1}{\frac{1}{2}}=-2

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

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

    y−(2)=−2(x−(4))y-(2)=-2(x-(4))

    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.

  6. Rearrange into the form y = mx + c

    y=−2x+10y=-2x+10

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

  7. Find the midpoint

    M=(6+32, 3+82)=(92, 112)M=\left(\frac{6+3}{2},\ \frac{3+8}{2}\right)=\left(\frac{9}{2},\ \frac{11}{2}\right)

    The midpoint sits exactly halfway between the two points, so we average the x-values and average the y-values. Adding and dividing by 2 finds the middle.

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

    mBC=y2−y1x2−x1=8−(3)3−(6)=−53m_{BC}=\frac{y_2-y_1}{x_2-x_1}=\frac{8-(3)}{3-(6)}=-\frac{5}{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.

  9. Find the perpendicular gradient

    m′=−1−53=35m'=-\frac{1}{-\frac{5}{3}}=\frac{3}{5}

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

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

    y−(112)=35(x−(92))y-(\frac{11}{2})=\frac{3}{5}(x-(\frac{9}{2}))

    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.

  11. Rearrange into the form y = mx + c

    y=35x+145y=\frac{3}{5}x+\frac{14}{5}

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

  12. Solve the two perpendicular bisectors together

    −2x+10=35x+145-2x+10=\frac{3}{5}x+\frac{14}{5}

    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=3613x=\frac{36}{13}

    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=5813y=\frac{58}{13}

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

  15. Verify it is equidistant from the vertices

    OA2=2125169OA^2=\frac{2125}{169}

    As a check, the distance to each vertex should be equal (the circumradius). This confirms the centre is correct.

  16. State the circumcentre

    (3613, 5813)\left(\frac{36}{13},\ \frac{58}{13}\right)

    This is the single point equally far from A, B and C, i.e. the centre of the circle through them.

Answer
(3613, 5813)\left(\frac{36}{13},\ \frac{58}{13}\right)
Question 5
8 markschallenging
A(1, 1)A\left(1,\ 1\right) and C(7, 5)C\left(7,\ 5\right) are opposite vertices of a rhombus ABCDABCD. Find the equation of the diagonal BDBD, in the form y=mx+cy=mx+c.
Show worked solution

Worked solution

  1. Understand the problem

    find the line containing diagonal BD\text{find the line containing diagonal }BD

    We know one diagonal AC of the rhombus. We must find the equation of the line through the other diagonal BD.

  2. Recall the diagonal property of a rhombus

    diagonals bisect each other at right angles\text{diagonals bisect each other at right angles}

    In a rhombus the diagonals cross at their midpoint and meet at 90 degrees. So BD is the perpendicular bisector of AC.

  3. Write down the given vertices

    A(1, 1), C(7, 5)A\left(1,\ 1\right),\ C\left(7,\ 5\right)

    These are the opposite corners joined by diagonal AC. BD must pass through the middle of AC.

  4. Find the midpoint of AC

    M=(1+72, 1+52)=(4, 3)M=\left(\frac{1+7}{2},\ \frac{1+5}{2}\right)=\left(4,\ 3\right)

    Average the two opposite vertices to find the crossing point of the diagonals.

  5. Find the length of AC (context)

    AC=213AC=2\sqrt{13}

    The length of the known diagonal is useful context; the diagonals of a rhombus generally differ in length.

  6. Find the gradient of AC

    mAC=5−(1)7−(1)=23m_{AC}=\frac{5-(1)}{7-(1)}=\frac{2}{3}

    We need this to make the other diagonal perpendicular to AC.

  7. Find the perpendicular gradient

    m′=−123=−32m'=-\frac{1}{\frac{2}{3}}=-\frac{3}{2}

    Take the negative reciprocal for the second diagonal BD.

  8. Form the equation of diagonal BD

    y−(3)=−32(x−(4))y-(3)=-\frac{3}{2}(x-(4))

    Use the midpoint and perpendicular gradient in point-gradient form.

  9. Multiply through to clear the fraction

    2(y−(3))=−3(x−(4))2\big(y-(3)\big)=-3\big(x-(4)\big)

    Clearing the fraction avoids awkward arithmetic and keeps the working tidy.

  10. Expand both sides

    2y−6=−3x+122y-6=-3x+12

    Multiply out each bracket carefully, watching signs.

  11. Rearrange to y = mx + c

    y=−32x+9y=-\frac{3}{2}x+9

    Make y the subject to give the equation of the diagonal BD.

  12. Check M lies on the line

    −32(4)+9=3-\frac{3}{2}(4)+9=3

    Substituting the midpoint M gives its y-coordinate, confirming BD passes through the centre.

  13. Check BD is perpendicular to AC

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

    The product of the two diagonal gradients is -1, so the diagonals meet at right angles as a rhombus requires.

  14. Interpret the result

    B, D lie on this line, symmetric about MB,\ D\ \text{lie on this line, symmetric about }M

    The two remaining vertices B and D sit on this line, one on each side of M and equally far from it.

  15. State the equation of diagonal BD

    y=−32x+9y=-\frac{3}{2}x+9

    This perpendicular bisector of AC is the line containing the other diagonal of the rhombus.

Answer
y=−32x+9y=-\frac{3}{2}x+9

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