Perpendiculars: Feet and Areas

Foot of the perpendicular: build the perpendicular, solve with the original line

In the exam for Edexcel International AS Maths · 5 real past-paper questions below

The Foot of the Perpendicular

You can already find a line perpendicular to another. AS questions go one step further: they ask where that perpendicular meets the first line.

That meeting point is called the foot of the perpendicular. It is the point on the line closest to your point.

Steps

1
Find the gradient of the given line, then flip it and change the sign.
2
Write the perpendicular through your point.
3
Solve the two equations together. The answer is the foot.

Worked example

1
Find the foot of the perpendicular from $A (5, - 1)$ to the line $l : y = \frac{1}{2} x + 4$ .
2
$m_{l} = \frac{1}{2}$ — The number in front of x in l.
3
$m = - 2$ — Flip and change the sign. Check: $\frac{1}{2} \times (- 2) = - 1$ .
4
$y - (- 1) = - 2 (x - 5)$ — The perpendicular passes through A(5, −1).
5
$y = - 2 x + 9$ — Multiply out: $y + 1 = - 2 x + 10$ , then take 1 from both sides.
6
$\frac{1}{2} x + 4 = - 2 x + 9$ — Both lines have the same y where they meet, so set them equal.
7
$\frac{5}{2} x = 5$ — Add 2x to both sides and take 4 from both sides.
8
$x = 2$ — Multiply both sides by $\frac{2}{5}$ .
9
$y = - 2 (2) + 9 = 5$ — Put x = 2 into either line. The foot is $F (2, 5)$ .

Common mistake

Solving the perpendicular with the wrong line. The foot lies on l, so solve the perpendicular together with l itself.

Rule

These questions often say solutions relying on calculator technology are not acceptable. Show the line where you collect the x terms, like $\frac{5}{2} x = 5$ . An answer that just appears loses the method mark.

Key idea

Check: F must fit both lines. $\frac{1}{2} (2) + 4 = 5$ and $- 2 (2) + 9 = 5$ . It does.

Exact Area With a Perpendicular Height

Once you have the foot, you have a height. The base of a triangle lies along the line, and the height is the perpendicular from the third corner.

Formula

Area = \frac{1}{2} \times P Q \times A F

$P Q$ — the base: the length between two points on the line.
$A F$ — the height: from the corner A to the foot F. It must meet the line at a right angle.

Worked example

1
P(−4, 2) and Q(6, 7) lie on $l : y = \frac{1}{2} x + 4$ . Using F(2, 5) from before, find the exact area of triangle APQ.
2
$P Q^{2} = (6 - (- 4))^{2} + (7 - 2)^{2}$ — Distance between two points: Pythagoras on the across and up.
3
$P Q = 125 = 55$ — $100 + 25 = 125$ and $125 = 255$ .
4
$A F^{2} = (5 - 2)^{2} + (- 1 - 5)^{2}$ — A is (5, −1) and F is (2, 5).
5
$A F = 45 = 35$ — $9 + 36 = 45$ and $45 = 95$ .
6
$\frac{1}{2} \times 55 \times 35$ — Base times height, halved.
7
$= \frac{1}{2} \times 15 \times 5 = \frac{75}{2}$ — $5 \times 5 = 5$ . The surds vanish and the answer is exact.

Common mistake

Using AP or AQ as the height. They are slanted, not perpendicular, so the area comes out too big.

Rule

"Exact area" means keep the surds until they cancel. Decimals like 11.18 × 6.71 lose the accuracy mark.

Key idea

If the base lies on the x-axis, the height is easier: it is just the y-coordinate of the top corner. No foot needed.

Key idea

The perpendicular is the shortest way from a point to a line. Road planners use this to link a new house to the nearest point on a road.

A Point at a Given Area

Now the area is given and a corner is missing. The point C lies on the perpendicular bisector of AB, the line through the midpoint M at right angles to AB.

Here MC is the height and AB is the base. So the area tells you how far C is from M. C can be that far on either side of AB, which gives two answers.

Worked example

1
A is (−2, 1) and B is (6, 7). C lies on the perpendicular bisector of AB, and triangle ABC has area 50. Find both positions of C.
2
$M = (2, 4), m_{A B} = \frac{6}{8} = \frac{3}{4}$ — Midpoint: average the x's and the y's. Gradient: up 6, across 8.
3
$A B = 8^{2} + 6^{2} = 10$ — Pythagoras on the across and up.
4
$\frac{1}{2} \times 10 \times M C = 50$ — Base AB, height MC.
5
$M C = 10$ — Divide both sides by 5.
6
$m = - \frac{4}{3}$ — The bisector is perpendicular to AB: flip and change the sign.
7
$3 across, 4 down \Rightarrow 5$ — Gradient $- \frac{4}{3}$ means 3 across for every 4 down. By Pythagoras that move is 5 long.
8
$6 across, 8 down \Rightarrow 10$ — Double it to get the 10 you need.
9
$C = (2 + 6, 4 - 8) = (8, - 4)$ — Step one way from M.
10
$C = (2 - 6, 4 + 8) = (- 4, 12)$ — Step the other way.

Memory trick

Gradient $- \frac{4}{3}$ : think of a 3-4-5 triangle as a staircase step along the line. Count how many steps you need.

Common mistake

Giving only one C. The question says "two possible" for a reason: one on each side of AB. The last mark needs both.

Key idea

Check: both points fit the bisector $4 x + 3 y - 20 = 0$ . $32 - 12 - 20 = 0$ and $- 16 + 36 - 20 = 0$ .

Example

Two answers also appear when C lies on a level line and a length is given. B is (6, 4), C lies on $y = 1$ and $B C = 5$ .

$(x - 6)^{2} + (4 - 1)^{2} = 25$ , so $(x - 6)^{2} = 16$ and $x - 6 = \pm 4$ . So C is (10, 1) or (2, 1). Keep both signs.

Lines With an Unknown Constant

Some questions hide a constant, like k, in a curve or a line. You get the same line two ways, each with k in it. Then make them agree.

Worked example

1
The curve $y = \frac{1}{3} (x + 3) (k - x)$ , with $k > 0$ , cuts the x-axis at −3 and at k. Line $l_{1}$ has gradient $\frac{1}{3}$ and goes through (−3, 0). It meets the curve again at P. Line $l_{2}$ is perpendicular to $l_{1}$ , and passes through both (k, 0) and P. Find k.
2
$l_{1} : y = \frac{1}{3} (x + 3)$ — Gradient $\frac{1}{3}$ through (−3, 0).
3
$\frac{1}{3} (x + 3) (k - x) = \frac{1}{3} (x + 3)$ — Meet the curve: set them equal.
4
$k - x = 1 \Rightarrow x = k - 1$ — Divide by $\frac{1}{3} (x + 3)$ . That factor is zero only at the other point, x = −3.
5
$P = (k - 1, \frac{k + 2}{3})$ — Put $x = k - 1$ into $l_{1}$ : $\frac{1}{3} (k + 2)$ .
6
$l_{2} : y = - 3 (x - k) = - 3 x + 3 k$ — Way 1: gradient −3 (flip $\frac{1}{3}$ , change sign) through (k, 0).
7
$y - \frac{k + 2}{3} = - 3 (x - k + 1)$ — Way 2: gradient −3 through P.
8
$y = - 3 x + \frac{10 k - 7}{3}$ — Multiply out: $3 k - 3 + \frac{k + 2}{3} = \frac{9 k - 9 + k + 2}{3}$ .
9
$3 k = \frac{10 k - 7}{3}$ — Same line, same gradient, so the intercepts must match.
10
$9 k = 10 k - 7 \Rightarrow k = 7$ — Multiply both sides by 3, then take 9k from both sides.

Common mistake

Expanding $(x + 3) (k - x)$ into a quadratic. It works, but it is long. Spot the common factor $(x + 3)$ and cancel it.

Rule

If a part asks you to show that another equation for the line is true, write every line of the multiplying out. In the “hence” part, set the two forms equal.

Key idea

Check with k = 7: P is (6, 3). The curve gives $\frac{1}{3} (9) (1) = 3$ and $l_{2}$ gives $- 3 (6 - 7) = 3$ . Both agree.

Which method? Read the question

Each part tells you what it needs. Read it, then pick the tool.

You see	You do
"Perpendicular … meet at R"	Perpendicular through the point, then solve with the first line
"Exact area" of a triangle with a side on a line	½ × base along the line × perpendicular height; keep surds
A triangle with its base on the x-axis	½ × base × y-coordinate of the top corner
"C lies on l … area is …" with two answers	Area gives MC; step along l both ways from M
A length from a fixed point, two answers	Pythagoras; take ± of the square root
An unknown like k or b in a line or curve	Write the same line two ways; equate gradients or intercepts
"Hence" after a "show that"	Use the line you just showed; do not start again

Key idea

Many of these marks are method marks. A slip in one number still keeps them, as long as every line is shown.

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