Coordinate Geometry

8 methods · 17 questions

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When asked for the area enclosed by three lines where one is the $x$ -axis, $y$ -axis, or parallel to an axis, do not split the shape or use integration. Use the axis-aligned segment as the base.

Trigger: 'enclose a region with area' or 'area of the shape bounded by', involving $x = 0$ , $y = 0$ , or $x = c$ .

Instances:
(i) Triangle with a side on the $y$ -axis: the base length is the difference between the $y$ -intercepts of the other two lines.
(ii) The corresponding height is the absolute $x$ -coordinate of the intersection of those two lines.
(iii) Trapezium bounded by $x = a$ , $x = b$ , $y = 0$ and a slanted line: the parallel vertical sides are the heights of the slanted line at $a$ and $b$ , and the width is $∣ b - a ∣$ .

Linked questions (3)

NSAA 2019 Section 1, Part E (Mathematics questions) · Q73

Find the area of the shape bounded by the four lines:

2 y + x = 4

x = - 6

x = 0

y = 0

A $4$
B $12$
C $21$
D $25$
E $27$
F $30$

Worked solution

Answer:C

The two vertical lines

x = - 6

and

x = 0

cut off a strip of width

6

, and

y = 0

closes it below. The slanted line closes it above.

Write the slanted line as

y = \frac{4 - x}{2}

. At

x = - 6

its height is

\frac{4 + 6}{2} = 5

; at

x = 0

its height is

\frac{4}{2} = 2

. Both heights are positive, so the four lines really do bound a single region: a trapezium with two parallel vertical sides of lengths

5

and

2

, a distance

6

apart.

The area of a trapezium is the mean of the parallel sides times the distance between them:

\frac{1}{2} (5 + 2) \times 6 = 21.

The answer is C.

NSAA 2018 Section 1, Part A (Mathematics questions) · Q8

The straight lines

5 x + 2 y = 20

y = 3 x - 23

x = 0

enclose a region with area

K

square units.

What is the value of

K

A $39$
B $78$
C $99$
D $129$
E $198$
F $258$

Worked solution

Answer:C

The three lines meet in pairs at the three vertices of a triangle.

The line

5 x + 2 y = 20

meets

x = 0

(0, 10)

, and

y = 3 x - 23

meets

x = 0

(0, - 23)

.

The other two lines meet where

5 x + 2 (3 x - 23) = 20

, that is

11 x = 66

, so

x = 6

and

y = 3 (6) - 23 = - 5

: the third vertex is

(6, - 5)

.

Take the side lying on the

y

-axis as the base. Its length is

10 - (- 23) = 33

, and the corresponding height is the horizontal distance from

(6, - 5)

to the

y

-axis, namely

6

. Hence

K = \frac{1}{2} \times 33 \times 6 = 99.

The answer is C.

ENGAA 2020 Section 1, Part B (Mathematics questions) · Q29

The line L with equation

y = m x + c

, where

m > 0

and

c \geq 0

, passes through the point

(2, 4)

.

A line is drawn through the point

(2, 4)

perpendicular to L.

The triangle enclosed between the two lines and the

y

-axis has area 5 square units.

What is the larger of the two possible values of

m

A $- 0.5$
B $0.5$
C $1.25$
D $2$
E $5$

Worked solution

Answer:D

Since L passes through

(2, 4)

, we have

4 = 2 m + c

, so L meets the

y

-axis at

c = 4 - 2 m

.

The perpendicular through

(2, 4)

has gradient

- \frac{1}{m}

, so its equation is

y = - \frac{1}{m} (x - 2) + 4

and it meets the

y

-axis at

4 + \frac{2}{m}

.

The triangle has its apex at

(2, 4)

and its third side on the

y

-axis. That side has length

(4 + \frac{2}{m}) - (4 - 2 m) = 2 m + \frac{2}{m} = 2 m + \frac{2}{m}

because

m > 0

, and the height from

(2, 4)

to the

y

-axis is

2

. Setting the area equal to

5

\frac{1}{2} \times (2 m + \frac{2}{m}) \times 2 = 5 ⟹ 2 m + \frac{2}{m} = 5 ⟹ 2 m^{2} - 5 m + 2 = 0.

This factorises as

(2 m - 1) (m - 2) = 0

, so

m = \frac{1}{2}

m = 2

. Both are positive, and both give an admissible intercept

c = 4 - 2 m \geq 0

, namely

c = 3

and

c = 0

. The larger of the two values is

2

.

The answer is D.