Practice · HKDSE Mathematics Compulsory Part · Paper 2 multiple-choice

Revising this topic? Also work the Paper 1 long questions.

Equations of Straight Lines — Paper 2 Multiple Choice

Twenty original multiple-choice questions on coordinate geometry of the straight line, graded the way HKDSE Paper 2 grades them: routine slope and intercept work early, then loci, concurrency and area questions that carry an unknown constant. Answers are on the page. Detailed solutions can be requested by email.

20 questions · about 35 minutes · answers below each question

All questions are original, written for Math and AI Academy and modelled on past-paper style. They are not reproductions of HKEAA examination questions.

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Question 1 — equation of a line through two given points
Question 2 — parallel lines; y-intercept
Question 3 — perpendicular lines with the unknown in both equations
Question 4 — signs of the coefficients of a line read from a figure
Question 5 — mid-point formula used backwards
Question 6 — dividing a segment in a given ratio
Question 7 — reflection in y = x followed by a rotation about the origin
Question 8 — locus of a point equidistant from two fixed points
Question 9 — locus of a point equidistant from two parallel lines
Question 10 — x-intercepts of two parallel lines; distance between two points on an axis
Question 11 — perpendicular lines meeting on the y-axis; area used to recover a constant
Question 12 — three concurrent lines
Question 13 — line through the intersection of two lines and through the origin
Question 14 — reading a constant off a straight-line log graph
Question 15 — telling a parallel distinct line from an identical one
Question 16 — locating the intersection point by quadrant
Question 17 — perpendicular to the line joining two points; y-intercept
Question 18 — perpendicular lines meeting on an axis
Question 19 — minimising a total distance using a reflection
Question 20 — area of the triangle cut off by a line with an unknown slope

Question 1 · Modelled on 2024 P1 Q12

The straight line

L

passes through the points

(-2,\,5)

and

(6,\,-1)

. The equation of

L

$3x + 4y - 14 = 0$ .
$3x + 4y + 14 = 0$ .
$4x + 3y - 7 = 0$ .
$4x - 3y + 23 = 0$ .

Show answer

A. The slope is $\dfrac{-1-5}{6-(-2)} = -\dfrac{3}{4}$ , and substituting $(-2,\,5)$ gives $3x+4y-14=0$ . Option C swaps the two coefficients, which is the usual slip.

Question 2 · Modelled on 2021 P2 Q14

The straight line

L

passes through the point

(2,\,-1)

and is parallel to the straight line

5x - 2y + 7 = 0

. The

y

-intercept of

L

$-11$ .
$-6$ .
$4$ .
$6$ .

Show answer

B. Parallel means the same slope, $\dfrac{5}{2}$ . Then $y = \dfrac{5}{2}x - 6$ , so the $y$ -intercept is $-6$ .

Question 3 · Modelled on 2019 P2 Q24

It is given that the straight lines

3x + ky + 1 = 0

and

(k-2)x - 6y + 5 = 0

are perpendicular to each other, where

k

is a constant. Find

k

$-6$
$-3$
$-2$
$2$

Show answer

C. The slopes are $-\dfrac{3}{k}$ and $\dfrac{k-2}{6}$ . Setting their product to $-1$ gives $-3(k-2) = -6k$ , so $k = -2$ .

Question 4 · Modelled on 2019 P2 Q23 and 2026 P2 Q14

In the figure, the equation of the straight line

L

ax + by + c = 0

. Which of the following are true?

I.

ab < 0

II.

ac < 0

III.

bc > 0

I and II only
I and III only
II and III only
I, II and III

Show answer

D. The slope $-\dfrac{a}{b}$ is positive, so $ab < 0$ . The $x$ -intercept $-\dfrac{c}{a}$ is positive, so $ac < 0$ . The $y$ -intercept $-\dfrac{c}{b}$ is negative, so $bc > 0$ . No individual sign can be determined, only the three products.

Question 5 · Modelled on 2023 P2 Q23

M

is the mid-point of

PQ

. The coordinates of

P

and

M

are

(-3,\,7)

and

(1,\,2)

respectively. The coordinates of

Q

are

$\left(-1,\,\dfrac{9}{2}\right)$ .
$(5,\,-3)$ .
$(5,\,3)$ .
$(-7,\,12)$ .

Show answer

B. Each coordinate of $M$ is the average of the two endpoints, so $x_Q = 2(1) - (-3) = 5$ and $y_Q = 2(2) - 7 = -3$ . Option D reverses the direction.

Question 6 · Modelled on 2021 P2 Q25

The coordinates of the points

A

and

B

are

(-4,\,1)

and

(6,\,11)

respectively.

R

is a point lying on

AB

such that

AR : RB = 2 : 3

. The coordinates of

R

are

$(0,\,5)$ .
$(2,\,7)$ .
$(-2,\,3)$ .
$(1,\,6)$ .

Show answer

A. $R = \left(\dfrac{3(-4)+2(6)}{5},\ \dfrac{3(1)+2(11)}{5}\right) = (0,\,5)$ . Option B comes from using the ratio the wrong way round.

Question 7 · Modelled on 2024 P2 Q24

A

is the point

(3,\,-5)

A

is reflected about the straight line

y = x

A'

A'

is then rotated clockwise about the origin through

90^\circ

A''

. The coordinates of

A''

are

$(-3,\,-5)$ .
$(-5,\,-3)$ .
$(5,\,3)$ .
$(3,\,5)$ .

Show answer

D. Reflection in $y = x$ swaps the coordinates, giving $A'(-5,\,3)$ . A clockwise rotation of $90^\circ$ sends $(x,\,y)$ to $(y,\,-x)$ , giving $A''(3,\,5)$ .

Question 8 · Modelled on 2023 P2 Q26

The coordinates of the points

A

and

B

are

(2,\,-3)

and

(8,\,5)

respectively.

P

is a moving point such that

AP = BP

. The equation of the locus of

P

$4x - 3y - 17 = 0$ .
$3x + 4y + 19 = 0$ .
$3x + 4y - 19 = 0$ .
$4x + 3y - 23 = 0$ .

Show answer

C. The locus is the perpendicular bisector of $AB$ : it passes through the mid-point $(5,\,1)$ with slope $-\dfrac{3}{4}$ . Option A uses the slope of $AB$ itself.

Question 9 · Modelled on 2026 P2 Q26

The equations of the straight lines

L_1

and

L_2

are

5x - 12y + 26 = 0

and

5x - 12y - 52 = 0

respectively.

P

is a moving point in the rectangular coordinate plane such that the perpendicular distance from

P

L_1

is equal to the perpendicular distance from

P

L_2

. The locus of

P

a point.
a circle.
a straight line.
a pair of straight lines.

Show answer

C. $L_1$ and $L_2$ have the same slope and different intercepts, so they are parallel and never meet. The locus is the single line midway between them, $5x - 12y - 13 = 0$ . Option D would be right only if the two lines crossed.

Question 10 · Modelled on 2026 P2 Q15

The straight lines

4x - 3y + 12 = 0

and

4x - 3y - 8 = 0

cut the

x

-axis at

A

and

B

respectively. Find

AB

$4$
$5$
$20$
$25$

Show answer

B. Putting $y = 0$ gives $A(-3,\,0)$ and $B(2,\,0)$ , so $AB = 5$ . Option C adds the constants instead of the intercepts.

Question 11 · Modelled on 2018 P2 Q26

The straight line

L_1 : 3x - 4y + k = 0

, where

k

is a positive constant, cuts the

y

-axis at

C

. The straight line

L_2

is perpendicular to

L_1

and passes through

C

L_1

and

L_2

cut the

x

-axis at

A

and

B

respectively. If the area of

\triangle ABC

150

, then

k =

$24$ .
$36$ .
$48$ .
$60$ .

Show answer

C. $C\left(0,\,\dfrac{k}{4}\right)$ , $A\left(-\dfrac{k}{3},\,0\right)$ and $B\left(\dfrac{3k}{16},\,0\right)$ , so $AB = \dfrac{25k}{48}$ and the area is $\dfrac{25k^2}{384}$ . Setting this to $150$ gives $k = 48$ .

Question 12 · Modelled on 2022 P2 Q18

It is given that the straight lines

x + y - 6 = 0

2x - y - 3 = 0

and

3x + ky - 15 = 0

are concurrent, where

k

is a constant. Find

k

Show answer

A. The first two lines meet at $(3,\,3)$ . Substituting gives $9 + 3k - 15 = 0$ , so $k = 2$ .

Question 13 · Modelled on 2019 P2 Q20

The straight line

L

passes through the origin and through the point of intersection of the straight lines

x - 2y + 1 = 0

and

3x + y - 11 = 0

. The equation of

L

$2x + 3y = 0$ .
$3x + 2y = 0$ .
$3x - 2y = 0$ .
$2x - 3y = 0$ .

Show answer

D. The two lines meet at $(3,\,2)$ , so $L$ has slope $\dfrac{2}{3}$ and passes through the origin, giving $2x - 3y = 0$ . Option C inverts the slope.

Question 14 · Modelled on 2018 P2 Q33 and 2024 P1 Q15

It is given that

y = ab^x

, where

a

and

b

are positive constants. The graph of

\log y

against

x

is a straight line with slope

2

and

y

-intercept

3

. Find

b

$2$
$3$
$100$
$1000$

Show answer

C. Taking logarithms gives $\log y = (\log b)x + \log a$ . So $\log b = 2$ and $b = 100$ . Option D is $a$ , not $b$ .

Question 15 · Modelled on 2019 P2 Q17

Which of the following straight lines has no point of intersection with the straight line

6x - 9y + 4 = 0

$2x - 3y + 5 = 0$
$6x + 9y + 4 = 0$
$9x - 6y + 4 = 0$
$12x - 18y + 8 = 0$

Show answer

A. Option A has the same slope $\dfrac{2}{3}$ but a different intercept, so the two lines never meet. Option D is the same line written twice, so it meets the given line at every point.

Question 16 · Modelled on 2022 P2 Q24

The straight lines

3x + 2y + 12 = 0

and

x - 4y + 10 = 0

intersect at

P

P

lies in quadrant

I.
II.
III.
IV.

Show answer

B. Solving gives $P\left(-\dfrac{34}{7},\ \dfrac{9}{7}\right)$ . The $x$ -coordinate is negative and the $y$ -coordinate is positive, so $P$ is in quadrant II.

Question 17 · Modelled on 2025 P2 Q21

The straight line

L

passes through the point

(4,\,-2)

and is perpendicular to the straight line joining

(1,\,3)

and

(7,\,7)

. The

y

-intercept of

L

$-8$ .
$4$ .
$6$ .
$8$ .

Show answer

B. The joining line has slope $\dfrac{2}{3}$ , so $L$ has slope $-\dfrac{3}{2}$ . Then $y = -\dfrac{3}{2}x + 4$ .

Question 18 · Modelled on 2018 P2 Q25

The straight line

L_2

is perpendicular to the straight line

L_1 : 2x - 5y + 20 = 0

and intersects

L_1

at a point lying on the

y

-axis. The equation of

L_2

$5x + 2y - 8 = 0$ .
$5x + 2y - 20 = 0$ .
$2x + 5y - 20 = 0$ .
$5x - 2y + 8 = 0$ .

Show answer

A. $L_1$ cuts the $y$ -axis at $(0,\,4)$ , and $L_2$ has slope $-\dfrac{5}{2}$ , giving $5x + 2y - 8 = 0$ . Option B uses the wrong intercept.

Question 19 · Modelled on 2025 P2 Q25

The coordinates of the points

A

and

B

are

(1,\,3)

and

(9,\,5)

respectively.

P

is a point lying on the

x

-axis such that

AP + PB

is the least. The

x

-coordinate of

P

$2$ .
$3$ .
$\dfrac{10}{3}$ .
$4$ .

Show answer

D. Reflect $A$ in the $x$ -axis to $A'(1,\,-3)$ . Then $AP + PB = A'P + PB$ , which is least when $P$ lies on $A'B$ . The line $A'B$ is $y = x - 4$ , which cuts the $x$ -axis at $x = 4$ .

Question 20 · Modelled on 2023 P2 Q10

The straight line

L : y = mx + 4

, where

m < 0

, cuts the

x

-axis at

A

and cuts the

y

-axis at

B

. Denote the origin by

O

. If the area of

\triangle OAB

32

, then

m =

$-2$ .
$-1$ .
$-\dfrac{1}{2}$ .
$-\dfrac{1}{4}$ .

Show answer

D. $B(0,\,4)$ and $A\left(-\dfrac{4}{m},\,0\right)$ . Since $m < 0$ , $OA = -\dfrac{4}{m}$ and the area is $-\dfrac{8}{m} = 32$ , giving $m = -\dfrac{1}{4}$ .

Revising this topic? Also work the Paper 1 long questions.

Where this sits in the syllabus

Equations of straight lines sits in the Measures, Shape and Space strand. In Paper 2 it reliably supplies four to six questions a year, usually clustered near the end of Section A. The ones that cost marks are rarely the hard ones: they are the questions where a sign, a ratio order or a length written as a negative number quietly changes the answer.

Full coverage of this topic, and the rest of the course, is on the Mathematics page. More sets are listed on the practice index.

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