Linear Law

Important Phrase:

𝐴 against 𝐵 means that: 𝐴: Vertical axis

𝐵: Horizontal axis.

In other words, we always have: (𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑥𝑖𝑠) against (𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑥𝑖𝑠)

Idea:

This chapter is mainly on converting a curve (on the 𝑦 − 𝑥 axis) being converted to a straight

line (by “changing” the variables on the 𝑦 − 𝑥 axis) by a rearrangement (the so-call “change

of variables on the 2 axes) of the variables 𝒚 and 𝒙 in an non-linear equation relating 𝒚

and 𝒙 into the form of 𝒀 = 𝒎𝑿 + 𝒄 [where 𝒎 (gradient) and 𝒄 (𝒚-intercept) are constants].

where 𝑔(𝑥, 𝑦) and 𝑓(𝑥, 𝑦) are expressions involving variables 𝑥 and/or 𝑦. This

“rearrangement” of variables in the equation to form 𝑌 and 𝑋 (the new axes) is known as

“linearization” of the equation.

The new Vertical axis is named 𝑌

The new horizontal axis is named 𝑋

Do not mix up 𝑋 as 𝑥 and 𝑌 as 𝑦

Simple Example:

Given the equation: 𝑦 = 2𝑥2 + 1, how should the linearization be done?

The Linearization: 𝑦 = 2𝑥2 + 1

𝑦

𝑥

Becomes

𝑔(𝑥, 𝑦) [𝑌]

𝑓(𝑥, 𝑦) [𝑋]

Default axes New “Renamed” axes

𝑌 𝑋

𝑦

𝑥

Linearization

𝑦

𝑥2

Linearized Equation: 𝑌 = 2𝑋 + 1

𝑌 = 2𝑋 + 1

𝑦 = 2𝑥2 + 1

[𝑋]

[𝑌]

At times, algebraic manipulation will be needed in order to linearize the equation.

Here is an important rule:

The renamed axis 𝑌 and 𝑋 cannot consists of unknown constants.

Get the hang of the idea! Don’t just memorize the standard linearizations!

The following examples will illustrate the idea further.

Example:

Linearize the following non-linear equations:

(i) 𝑦 = 𝑎√𝑥 +𝑏

√𝑥

(ii) 𝑥 = 𝑏𝑥𝑦 + 𝑎𝑦

(iii) 𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

(iv) 𝑦 =𝑎

𝑥−𝑏

(v) 𝑦 = 𝑎𝑥𝑏

(vi) 𝑦 = 𝑎𝑏𝑥

(vii) 𝑦 = 𝑎𝑥𝑏 + 𝑐

(viii) 𝑦𝑎𝑥 = 𝑏 + 𝑐

(ix) 𝑦𝑏 = 10𝑥+𝑎

(x) 𝑦𝑏 = 7𝑥+𝑎

where 𝑎 and 𝑏 are unknown constants.

Solution:

(i) 𝑦 = 𝑎√𝑥 +𝑏

√𝑥

𝑦√𝑥 = 𝑎𝑥 + 𝑏

(ii) 𝑥 = 𝑏𝑥𝑦 + 𝑎𝑦

There are a couple of methods here.

Method 1: Method 2:

𝑥 = 𝑏𝑥𝑦 + 𝑎𝑦 𝑥 = 𝑏𝑥𝑦 + 𝑎𝑦

1 = 𝑏𝑥 +𝑎𝑦

𝑥

𝑥

𝑦= 𝑏𝑥 + 𝑎

𝑎𝑦

𝑥= −𝑏𝑥 + 1 [Choose 𝑌 =

𝑦

𝑥]

At times, there may

be more than 1 way

to linearize the

equations

Strategy to Linearize the Equations:

Try to ensure that there are only 2

terms (or 2 group of terms) with the

variables.

Write terms like 𝑏𝑥

𝑎 as (

𝑏

𝑎) 𝑥 [Write

the product of constant and variables

with separate brackets]

Ensure coefficient of 𝑌 is 1 so that it is

of the form 𝑌 = 𝑚𝑋 + 𝑐

When needing to apply logarithm on

both sides of the equation, when the

base is "𝑒", apply "𝑙𝑛" . When the base

is "10", apply “𝑙𝑔”. If not, it does not

matter whether we apply “ln” or “lg”

× √𝑥

𝑌 𝑋

𝑌 = 𝑦√𝑥

𝑋 = 𝑥

𝑚 = 𝑎

𝑐 = 𝑏

÷ 𝑥 ÷ 𝑦

𝑌 𝑋

𝑌 =𝑥

𝑦 , 𝑋 = 𝑥

𝑚 = 𝑏 , 𝑐 = 𝑎

𝑦

𝑥= (−

𝑏

𝑎) 𝑥 +

1

𝑎 Method 3:

𝑥 = 𝑏𝑥𝑦 + 𝑎𝑦

1

𝑦= 𝑏 + 𝑎 (

1

𝑥)

OR

1 = 𝑏𝑥 +𝑎𝑦

𝑥

𝑏𝑥 = −𝑎𝑦

𝑥+ 1 [Choose 𝑌 = 𝑥]

𝑥 = (−𝑎

𝑏) (

𝑦

𝑥) +

1

𝑏

Method 2 and 3 leads to shorter algebraic manipulation since the coefficient of one of the

variables is 1 after the division.

(iii)

𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

Take note that 𝑐 is not an unknown constant. Therefore it is legit to group it under 𝑋 or 𝑌.

𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐 ⟹ 𝑦 − 𝑐 = 𝑎𝑥2 + 𝑏𝑥

⟹ 𝑦 − 𝑐 = 𝑥(𝑎𝑥 + 𝑏)

⟹ 𝑦−𝑐

𝑥= 𝑎𝑥 + 𝑏

(iv)

𝑦 =𝑎

𝑥−𝑏 ⟹ 𝑦(𝑥 − 𝑏) = 𝑎 ⟹ 𝑦𝑥 − 𝑦𝑏 = 𝑎 ⟹ 𝑥𝑦 = 𝑏𝑦 + 𝑎

OR

𝑦 =𝑎

𝑥−𝑏 ⟹

1

𝑦=

𝑥−𝑏

𝑎 ⟹

1

𝑦=

𝑥

𝑎−

𝑏

𝑎⟹

1

𝑦= (

1

𝑎) 𝑥 + (−

𝑏

𝑎)

𝑌 𝑋

𝑌 =𝑦

𝑥 , 𝑋 = 𝑥

𝑚 = −𝑏

𝑎 , 𝑐 =

1

𝑎

÷ 𝑏

𝑌 𝑋

𝑌 = 𝑥 , 𝑋 =𝑦

𝑥

𝑚 = −𝑎

𝑏 , 𝑐 =

1

𝑏

÷ 𝑥𝑦

𝑌 𝑋 𝑌 =

1

𝑦 , 𝑋 =

1

𝑥

𝑚 = 𝑎 , 𝑐 = 𝑏

÷ 𝑥

𝑌 𝑋

𝑌 =𝑦−𝑐

𝑥 , 𝑋 = 𝑥

𝑚 = 𝑎 , 𝑐 = 𝑏

Cannot separate the unknown

constants and variables, so got

to get rid of fractions first!

𝑌 𝑋

𝑌 = 𝑥𝑦 , 𝑋 = 𝑦

𝑚 = 𝑏 , 𝑐 = 𝑎

𝑌 𝑋 𝑐

𝑌 =1

𝑦 , 𝑋 = 𝑥

𝑚 =1

𝑎 , 𝑐 = −

𝑏

𝑎

(v) 𝑦 = 𝑎𝑥𝑏

⟹ ln 𝑦 = ln(𝑎𝑥𝑏) = ln 𝑎 + ln(𝑥𝑏)

ln 𝑦 = ln 𝑎 + 𝑏 ln 𝑥

(vi) 𝑦 = 𝑎𝑏𝑥

⟹ ln 𝑦 = ln(𝑎𝑏𝑥)

⟹ ln 𝑦 = ln 𝑎 + ln(𝑏𝑥)

ln 𝑦 = ln 𝑎 + 𝑥(ln 𝑏)

(vii) 𝑦 = 𝑎𝑥𝑏 + 𝑐

While the logic to “separate” the unknown constants from the variable 𝑥 is the same (i.e. to

apply logarithms on both sides of the equation), one cannot apply log directly first, since

there is no way one can simplify ln(𝑎𝑥𝑏 + 𝑐) any further.

The key to proceed is that while we cannot simplify ln(𝑎𝑥𝑏 + 𝑐), we can simplify ln(𝑎𝑥𝑏):

𝑦 = 𝑎𝑥𝑏 + 𝑐 ⟹ 𝑦 − 𝑐 = 𝑎𝑥𝑏

⟹ ln(𝑦 − 𝑐) = ln(𝑎𝑥𝑏)

⟹ ln(𝑦 − 𝑐) = ln 𝑎 + ln(𝑥𝑏)

ln(𝑦 − 𝑐) = ln 𝑎 + 𝑏 ln 𝑥

(viii) 𝑦𝑎𝑥 = 𝑏 + 𝑐

ln(𝑦𝑎𝑥) = ln(𝑏 + 𝑐)

ln 𝑦 + ln(𝑎𝑥) = ln(𝑏 + 𝑐)

ln 𝑦 + 𝑥 ln 𝑎 = ln(𝑏 + 𝑐) ⟹ ln 𝑦 = (− ln 𝑎)𝑥 + ln(𝑏 + 𝑐)

Since 𝑏 (which is an unknown constant) is part of

the power of the variable 𝑥, we can only “separate”

𝑏 from 𝑥 by applying log on both sides.

Take Note:

ln(𝑎𝑥𝑏) ≠ 𝑏 ln(𝑎𝑥)

since the power 𝑏 only applies to 𝑥 and not to 𝑎

𝑌 𝑋

𝑌 = ln 𝑦 , 𝑋 = ln 𝑥

𝑚 = 𝑏 , 𝑐 = ln 𝑎

Since the base is 𝑥, it does

not matter whether I apply

“ln” or “lg”

Likewise, 𝑥 and 𝑏 can only be “separate” by means

of applying logarithms on both sides of the equation

𝑌 𝑋

𝑌 = ln 𝑦 , 𝑋 = 𝑥

𝑚 = ln 𝑏 , 𝑐 = ln 𝑎

𝑚 𝑐

ln(𝑎𝑥𝑏 + 𝑐) ≠ ln(𝑎𝑥𝑏) + ln 𝑐

Note that 𝑐 is NOT an unknown

constant (it is a known

constant), thus, 𝑐 can be group

with 𝑌 or 𝑋

𝑌 𝑋

𝑌 = ln(𝑦 − 𝑐) , 𝑋 = ln 𝑥

𝑚 = 𝑏 , 𝑐 = ln 𝑎 𝑚 𝑐

𝑌 𝑋 𝑚 𝑐

𝑌 = ln 𝑦 , 𝑋 = 𝑥

𝑚 = − ln 𝑎 , 𝑐 = ln(𝑏 + 𝑐)

(ix) 𝑦𝑏 = 10𝑥+𝑎

lg(𝑦𝑏) = lg(10𝑥+𝑎) ⟹ 𝑏 lg 𝑦 = (𝑥 + 𝑎)

2 possible ways of naming 𝑋 and 𝑌:

Way 1: Way 2:

𝑏 lg 𝑦 = (𝑥 + 𝑎) 𝑏 lg 𝑦 = (𝑥 + 𝑎)

𝑥 = 𝑏 lg 𝑦 − 𝑎 lg 𝑦 =1

𝑏(𝑥 + 𝑎)

lg 𝑦 = (1

𝑏) 𝑥 + (

𝑎

𝑏)

(x) 𝑦𝑏 = 7𝑥+𝑎

ln(𝑦𝑏) = ln(7𝑥+𝑎)

[Since the base is “7” (not “e” or “10”, it does not matter whether we apply “log” or “ln”]

𝑏 ln(𝑦) = (𝑥 + 𝑎) ln 7

𝑏 ln 𝑦 = 𝑥 ln 7 + 𝑎 ln 7

ln 𝑦 = (ln 7

𝑏) 𝑥 +

𝑎 ln 7

𝑏

Question on linear law often requires us to find the value of “unknown” constants based on

some given information.

𝑌 𝑋

𝑌 = 𝑥 , 𝑋 = lg 𝑦

𝑚 = 𝑏 , 𝑐 = −𝑎

𝑌 𝑋

𝑌 = lg 𝑦 , 𝑋 = 𝑥

𝑚 =1

𝑏 , 𝑐 =

𝑎

𝑏

÷ 𝑏

𝑌 𝑋

𝑌 = ln 𝑦 , 𝑋 = 𝑥

𝑚 =ln 7

𝑏 , 𝑐 =

𝑎 ln 7

𝑏

Example: [Expressing 𝑦 in terms of 𝑥]

The diagram shows part of a straight line obtained by plotting 𝑦+𝑥

𝑥 against 𝑥. Express 𝑦 in

terms of 𝑥 given that 𝑥 >1

2.

Solution:

From the axis seen:

𝑌 =𝑦+𝑥

𝑥 and 𝑋 = 𝑥

Equation of the line: 𝑌 = 𝑚𝑋 + 𝐶 --- (1)

𝑚 = 5−3

5−1=

1

2

Substitute (1,3) into Equation (1) [We can also substitute (5,5), it does not matter]

3 =1

2(1) + 𝐶 ⟹ 𝐶 =

5

2

∴ 𝑌 =1

2𝑋 +

5

2 --- (2)

Since we wanted a relation between 𝑦 and 𝑥 (Not 𝑌 and 𝑋), we substitute 𝑌 =𝑦+𝑥

𝑥 and

𝑋 = 𝑥 into (2):

𝑦+𝑥

𝑥=

1

2(𝑥) +

5

2 ⟹ 2(𝑦 + 𝑥) = 𝑥2 + 𝑥

2𝑦 + 2𝑥 = 𝑥2 + 𝑥 ⟹ 2𝑦 = 𝑥2 − 𝑥

2𝑦 = (𝑥 −1

2)

2

− (1

2)

2

(𝑥 −1

2)

2

= 2𝑦 +1

4 ⟹ 𝑥 −

1

2= ± √2𝑦 +

1

4

𝑥 =1

2± √2𝑦 +

1

4

Since 𝑥 >1

2, we must have: 𝑥 =

1

2+ √2𝑦 +

1

4

𝑦+𝑥

𝑥

𝑥 (1,3)

(5,5)

× 2𝑥

[Complete the square on the

RHS to make 𝑥 the subject]

Example: [Expressing 𝑦 in terms of 𝑥]

The diagram shows part of a straight line graph obtained by plotting ln 𝑦 against 𝑥3 and the

coordinates of two of the points on the line. Express 𝑦 in terms of 𝑥.

Solution:

From the axis seen:

𝑌 = ln 𝑦 and 𝑋= 𝑥3

Equation of the line: 𝑌 = 𝑚𝑋 + 𝐶 --- (1)

𝑚 = 13−4

5−2= 3

Substitute (2,4) into Equation (1):

4 = 3(2) + 𝐶 ⟹ 𝐶 = −2

∴ 𝑌 = 3𝑋 − 2 --- (2)

Now, substitute 𝑌 = ln 𝑦 and 𝑋 = 𝑥3 into (2):

ln 𝑦 = 3𝑥3 − 2

∴ 𝑦 = 𝑒3𝑥3−2

Example: [Finding value of unknown constant in the non-linear relation]

The variables 𝑥 and 𝑦 are connected by the equation 𝑦 = 𝑎𝑥𝑏 where 𝑎 and 𝑏 are constants.

The figure shows the straight line obtained by plotting lg 𝑦 against lg 𝑥. Calculate the value of

𝑎 and 𝑏 and hence, find the value of 𝑦 when 𝑥 = 4.

Solution:

From the given axes:

𝑌 = lg 𝑦 and 𝑋 = lg 𝑥

Equation of the line: 𝑌 = 𝑚𝑋 + 𝑐

𝑚 =0−(−1)

2−0=

1

2

Substitute (2,0) into Equation (1):

0 =1

2(2) + 𝐶 ⟹ 𝐶 = −1

ln 𝑦

𝑥3

(5,13)

(2,4)

Since log𝑒 𝑦 = 𝑎 ⟹ 𝑦 = 𝑒𝑎

lg 𝑦

lg 𝑥 (2,0)

(0, −1)

∴ 𝑌 =1

2𝑋 − 1 --- (2)

Now, substitute 𝑌 = lg 𝑦 and 𝑋 = lg 𝑥 into (2):

lg 𝑦 =1

2(lg 𝑥) − 1

lg 𝑦 = lg (𝑥1

2) − 1

lg 𝑦 = lg (𝑥1

2) − lg 10 ⟹ lg 𝑦 = lg (10𝑥1

2)

∴ 𝑦 = 10𝑥1

2

𝑎 = 10 and 𝑏 =1

2

Example:

The diagram shows part of the straight line drawn to represent the curve 𝑦 =𝑝𝑥

𝑥−𝑞, where 𝑝

and 𝑞 are constants. Find the value of 𝑝 and 𝑞.

Solution:

From the given axes:

𝑌 =1

𝑦 and 𝑋 =

1

𝑥

Equation of the line: 𝑌 = 𝑚𝑋 + 𝑐

𝑚 =7−1

2−5= −2

Substitute (2,7) into 𝑌 = 𝑚𝑋 + 𝑐:

7 = −2(2) + 𝐶 ⟹ 𝐶 = 11

∴ 𝑌 = −2𝑋 + 11

Since 𝑌 =1

𝑦 and 𝑋 =

1

𝑥

Compare this with the

given non-linear relation

𝑦 = 𝑎𝑥𝑏

1

𝑦

1

𝑥

(2,7)

(5,1)

1

𝑥

1

𝑦

∴ 1

𝑦=

2

𝑥+ 11 [Make 𝑦 the subject from here]

𝑥 = 2𝑦 + 11𝑥𝑦

𝑦(2 + 11𝑥) = 𝑥 ⟹ 𝑦 = 𝑥

11𝑥+2 [Algebraically re-write into the form of 𝑦 =

𝑝𝑥

𝑥−𝑞]

𝑥

11𝑥+2=

𝑥

11(𝑥+2

11)

=1

11𝑥

𝑥+2

11

∴ 𝑝 =1

11 and 𝑞 =

2

11

Practice:

Q1)

The table shows some of the experimental values of two variables 𝑥 and 𝑦.

𝑥 1.2 1.6 2.0 2.4 3.0

𝑦 3.23 3.58 3.90 4.19 4.57

It is known that 𝑥 and 𝑦 are related by the equation: 𝑦 = ln(𝐴𝑥2 − 𝐵) + 2.

(i) Obtain a straight line to represent the given data.

(ii) Use your graph to estimate the value of 𝐴 and of 𝐵.

(iii) By drawing a suitable straight line, find the value of 𝑥 and 𝑦 which satisfy the

simultaneous equations

𝑦 = ln(𝐴𝑥2 − 𝐵) + 2

𝑒𝑦 = 𝑒2(𝑥2 + 2)

[Ans: (ii) 𝐴 = 1.28, 𝐵 = −1.58 (iii) 𝑥 = 1.14, 𝑦 = 3.19]

Q2) [Linearization]

In order for each of the following equations:

(a) 𝑦 = 𝑒𝑎𝑥+𝑏

(b) 𝑝𝑦 + 𝑞𝑥 = 𝑥𝑦

(c) 𝐴𝑥2 + 𝐵𝑥 − 1 = 𝑥𝑦

(d) 𝑦 = 𝑎𝑏𝑥 − 10

× 𝑥𝑦

where 𝑎, 𝑏, 𝑝, 𝑞, 𝐴 and 𝐵 are unknown constants, to be expressed as a straight line, they need

to be expressed in the form of 𝑌 = 𝑚𝑋 + 𝑐, where 𝑋 and 𝑌 are functions of 𝑥 and/or 𝑦, and

𝑚 and 𝑐 are constants. Copy and complete the table for each of the following 𝑌, 𝑋, 𝑚 and 𝑐.

𝑌 𝑋 𝑚 𝑐

𝑦 = 𝑒𝑎𝑥+𝑏 𝑥

𝑝𝑦 + 𝑞𝑥 = 𝑥𝑦 1

𝑦

𝐴𝑥2 + 𝐵𝑥 − 1 = 𝑥𝑦 𝐵

𝑦 = 𝑎𝑏𝑥 − 10 𝑥

Q3)

The table shows some of the experimental values of two variables 𝑥 and 𝑦. It is known that 𝑥

and 𝑦 are related by the equation 𝑦 = 𝑎𝑥(𝑥2 + 𝑏), where 𝑎 and 𝑏 are constants, and that one

of the values of 𝑦 given in the table is subjected to an abnormally large error.

𝑥 1 1.5 2 2.5 3 3.5

𝑦 6.65 8.94 10 9.41 5.95 1.62

(a) On graph paper, plot 𝑦

𝑥 against 𝑥2, using a scale of 1 cm to represent 2 units on the 𝑥2

axis and 2 cm to represent 1 unit on the 𝑦

𝑥-axis. Draw a best fit line graph to represent

the equation of 𝑦 = 𝑎𝑥(𝑥2 + 𝑏).

(b) Use your graph to:

(i) Identify the abnormal reading and estimate its correct value.

(ii) Find the value of 𝑎 and 𝑏.

(iii) Estimate the value of 𝑥 when 𝑦 = 4𝑥.

(iv) State the gradient of the straight line if 𝑦

𝑥3 is plotted against 1

𝑥2.

[Ans: (b) (i) Abnormal reading: 𝑥2 = 9, correct value of 𝑦

𝑥= 2.25, correct value of 𝑦 = 6.75;

(b)(ii) 𝑎 = −0.53, 𝑏 = −13.5 (b)(iii) 2.41 (b)(iv) −7.155]

Q4)

The variables 𝑥 and 𝑦 are related by the equation 𝑦 = 10 + 10𝐴𝑏𝑥, where 𝐴 and 𝑏 are

constants. The table below shows experimental values of 𝑥 and 𝑦.

𝑥 10 15 20 25 30 35

𝑦 10.09 10.24 10.60 11.46 13.72 19.34

(i) Express this equation in a form suitable for drawing a straight line graph.

(ii) Draw this graph for the given data and use it to estimate 𝐴 and 𝑏.

(iii) Estimate the value of 𝑥 for which 𝑦 = 15.

(iv) By drawing a suitable straight line on your graph, solve the equation

(lg 𝑏 +1

20) 𝑥 + 𝐴 = 0

[Ans: (i) 𝑦 = lg(𝑦 − 10) = x(lg 𝑏) + 𝐴 (ii) 𝐴 = −1.82, 𝑏 = 1.20 (iii) 31.5 (iv) 𝑥 = 14.1]

Q5)

The diagram shows part of a straight line graph passing through the points 𝐴(4,6), 𝐵(𝑘, 10)

and 𝐶(12,18). Find:

(i) 𝑦 in terms of 𝑥,

(ii) the value of 𝑘,

(iii) the values of 𝑥 and 𝑦 at point 𝐴.

[Ans: (i) 𝑦 =3

2𝑥2 + 2𝑥 (ii)

20

3 (iii) (2,10) or (−2,2)]

Q6)

The figure shows part of a straight line graph obtained by plotting 𝑦(𝑥 + 2) against 𝑦. Given

that the variables 𝑥 and 𝑦 are related by the equation 𝑦 =4(3−𝑦)

2𝑥−1, find:

(i) The gradient of the straight line,

(ii) The value of 𝑐.

[Answer: (i) 1

2 (ii) 6]

𝑦 − 2𝑥

𝑥2 𝑂

𝐴(4,6)

𝐵(𝑘, 10)

𝐶(12,18)

𝑦(𝑥 + 2)

𝑦

𝑐

𝑂

Q7)

The line joining 𝐴(4,9) and 𝐵(0,2) meets the 𝑥-axis at 𝐶. The point 𝑃 lies on 𝐴𝐶 such that

AP: PC=2:3. A line through 𝑃 meets the 𝑥-axis at 𝑄 such that ∠𝑃𝐶𝑄 = ∠𝑃𝑄𝐶. Find:

(i) The coordinates of 𝑃,

(ii) The equation of 𝑃𝑄,

(iii) The value of 𝑥 and of 𝑦 at the point 𝑄.

[Answer: (i) (7

3,

24

5) (ii) 𝑦 = −

6

5𝑥2 +

38

5𝑥

(iii) 𝑥 =19

3, 𝑦 = 0]

𝑦

𝑥

𝑥 𝐶

𝐵(0,2)

𝐴(4,9) 𝑃

𝑂 𝑄