Shortest Distance from a Point to a Line

Find the shortest distance from a point to a line: it is the perpendicular, found with sine or by doubling the area and dividing by the base.

In the exam for Cambridge IGCSE Maths 0580, Cambridge O Level Maths 4024 · 19 real past-paper questions below

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Straight Across Is Shortest

From a point to a straight line there are many paths. The shortest one meets the line at a right angle. This is the perpendicular distance.

Rule

The shortest distance from a point to a line is the perpendicular. Any slanted path is longer, because it is the hypotenuse of a right triangle.

Key idea

"Shortest distance", "how close does it get" and "perpendicular height" all mean the same line.

Using Sine

You know how far the point is along a slanted line, and the angle that line makes. The distance you want is opposite the angle, and the slanted line is the hypotenuse.

A boat $P$ is 12 km from a harbour $A$ . The line $A P$ makes 35° with the straight coast. How close is $P$ to the coast?

Worked example

1
$sin 3 5^{\circ} = \frac{d}{12}$ — The shortest distance $d$ meets the coast at a right angle, so it is opposite the 35° angle. The slanted 12 km is the hypotenuse.
2
$d = 12 \times sin 3 5^{\circ}$ — Multiply both sides by 12 to get $d$ on its own.
3
$d = 12 \times 0.5736 = 6.883$ — Put $sin 3 5^{\circ}$ into your calculator, which must be in degrees mode.
4
$d = 6.88 km$ — Round to 3 significant figures and give the units.

Common mistake

Using cos: $12 cos 3 5^{\circ} = 9.83$ is how far along the coast, not the distance to it.

Sometimes the shortest distance is given and the slanted length is wanted. A boat $B$ is 7.5 km from a straight coast at its closest point. The line from $B$ to a harbour $H$ makes 28° with the coast. How far is $B$ from $H$ ?

Worked example

1
$sin 2 8^{\circ} = \frac{7.5}{B H}$ — The 7.5 km is the shortest distance, so it is opposite the 28° angle. The slanted line $B H$ is the hypotenuse.
2
$B H \times sin 2 8^{\circ} = 7.5$ — Multiply both sides by $B H$ to bring it out from the bottom of the fraction.
3
$B H = \frac{7.5}{s i n 2 8 ^{\circ}}$ — Divide both sides by $sin 2 8^{\circ}$ .
4
$B H = \frac{7.5}{0.46947} = 15.98$ — Work it out. Dividing by a number below 1 makes the answer bigger.
5
$B H = 16.0 km$ — Round to 3 significant figures.

Key idea

Check it makes sense. The slanted line is always longer than the shortest distance, and 16.0 is bigger than 7.5.

Using the Area

In a triangle, the perpendicular from a corner to the opposite side is the height. Find the area first, then work back.

Formula

d = \frac{2 \times Area}{base}

A triangle has sides 10 cm and 15 cm with 40° between them. Find the shortest distance from the far end of the 10 cm side to the 15 cm side.

Worked example

1
$Area = \frac{1}{2} \times 10 \times 15 \times sin 4 0^{\circ}$ — Two sides and the angle between them, so use area = ½ab sin C.
2
$Area = 75 \times 0.6428 = 48.21 cm^{2}$ — Half of 10 × 15 is 75. Then multiply by $sin 4 0^{\circ}$ .
3
$48.21 = \frac{1}{2} \times 15 \times d$ — The area is also ½ × base × height. Take the 15 cm side as the base, so the height is the shortest distance $d$ .
4
$d = \frac{2 \times 48.21}{15}$ — Multiply both sides by 2 to remove the half, then divide by the base 15.
5
$d = \frac{96.42}{15} = 6.43 cm$ — Work it out and round to 3 significant figures.

Key idea

Check with sine: $10 sin 4 0^{\circ} = 6.43$ . Both ways agree.

Common mistake

Forgetting to double the area. Dividing 48.21 by 15 gives 3.21, which is half the height.

Two Steps: Find a Side First

Often the angle you are given is not next to the distance you want. Find a side first, then use sine. In triangle $A B C$ , $A C = 18$ cm, angle $A = 3 5^{\circ}$ and angle $B = 10 0^{\circ}$ . Find the shortest distance from $B$ to $A C$ .

Worked example

1
$∠ C = 18 0^{\circ} - 3 5^{\circ} - 10 0^{\circ} = 4 5^{\circ}$ — Angles in a triangle add to 180°. Now all three angles are known.
2
$\frac{B C}{s i n 3 5 ^{\circ}} = \frac{18}{s i n 10 0 ^{\circ}}$ — Sine rule. The side 18 faces the 100° angle, and $B C$ faces the 35° angle.
3
$B C = \frac{18 s i n 3 5 ^{\circ}}{s i n 10 0 ^{\circ}} = 10.48$ — Multiply both sides by $sin 3 5^{\circ}$ . Keep 4 figures, because it is used again.
4
$B H ⊥ A C$ — Draw the shortest distance from $B$ straight down to $A C$ , meeting it at $H$ . Triangle $B H C$ has a right angle at $H$ .
5
$sin 4 5^{\circ} = \frac{B H}{10.48}$ — In triangle $B H C$ , $B H$ is opposite angle $C$ , and $B C$ is the hypotenuse.
6
$B H = 10.48 \times sin 4 5^{\circ} = 7.41 cm$ — Multiply both sides by 10.48 and round to 3 significant figures.

Key idea

Check it the other way. $A B = \frac{18 s i n 4 5 ^{\circ}}{s i n 10 0 ^{\circ}} = 12.92$ and $12.92 sin 3 5^{\circ} = 7.41$ . Both routes agree.

Common mistake

Stopping at $B C = 10.48$ . That is a slanted side of the triangle, not the shortest distance. Always finish with the perpendicular.

Which Method? Read the Question

Look for a right angle you can make.

You see	You do
Slanted length and its angle	distance = length × sin(angle)
Shortest distance given, slanted length wanted	length = distance ÷ sin(angle)
The angle is not next to the distance	find a side first, then multiply by sin
Area known, or easy to find	double the area, divide by the base
Two sides and the angle between	Area of a Triangle Using Sine
A right-angled triangle already	Choosing the Right Trigonometric Ratio

Key idea

Mark the right angle on your sketch first. Then the side you want is clear.

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