ONE EIGHTH CONTEST 2025
Ghana National College - 65 points
Tema Secondary School - 33 points
Effiduase SHCS - 28 points
ONE EIGHTH CONTEST 2025
Mankranso SHS - 43 points
St Paul's SHS, Denu - 23 points
Adonten SHS - 14 points
ONE EIGHTH CONTEST 2025
St. Thomas Aquinas SHS - 51 points
Tamale SHS - 37 points
O'Reilly SHS - 25 points
ROUND 2
COORDINATE GEOMETRY
QUESTION
Find the distance from the origin to the line 3x plus 4y plus 20 equals zero.
Find the distance from the origin $(0,0)$ to the line $3x+4y-20=0$.
ANSWER: $4\text{ units}$
SOLUTION 1
Rule: The shortest distance from the origin $(0,0)$ to the line $ax+by+c=0$ simplifies directly to:
$\text{Distance} = \dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=3, b=4, c=-20$:
$\text{Distance} = \dfrac{|-20|}{\sqrt{3^2+4^2}}$
$\text{Distance} = \dfrac{20}{\sqrt{25}}$
$\text{Distance} = \dfrac{20}{5}$
$\text{Distance} = 4\text{ units}$
SOLUTION 2
The distance from a general point $(h,k)$ to the line $ax+by+c=0$ is given by:
$\text{Distance} = \dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substituting the origin $(0,0)$ where $h=0$ and $k=0$:
$\text{Distance} = \dfrac{|3(0)+4(0)-20|}{\sqrt{3^2+4^2}}$
$\text{Distance} = \dfrac{|-20|}{\sqrt{9+16}}$
$\text{Distance} = \dfrac{20}{\sqrt{25}}$
$\text{Distance} = \dfrac{20}{5}$
$\text{Distance} = 4\text{ units}$
PRACTICE QUESTIONS
1. Find the distance from the origin to the line $-4x+3y=40$.
ANSWER: $8$ units
SOLUTION 1
Rule: the distance from the origin to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=-4$, $b=3$, $c=-40$ (writing the line as $ax+by+c=0$):
$\text{Distance}=\dfrac{|-40|}{\sqrt{(-4)^2+3^2}}$
$\text{Distance}=\dfrac{40}{\sqrt{25}}$
$\text{Distance}=\dfrac{40}{5}=8\text{ units}$
SOLUTION 2
The distance from a point $(h,k)$ to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substitute the origin $(0,0)$:
$\text{Distance}=\dfrac{|-4(0)+3(0)-40|}{\sqrt{(-4)^2+3^2}}$
$\text{Distance}=\dfrac{40}{\sqrt{25}}=8\text{ units}$
2. Find the distance from the origin to the line $-8x-6y=-20$.
ANSWER: $2$ units
SOLUTION 1
Rule: the distance from the origin to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=-8$, $b=-6$, $c=20$ (writing the line as $ax+by+c=0$):
$\text{Distance}=\dfrac{|20|}{\sqrt{(-8)^2+(-6)^2}}$
$\text{Distance}=\dfrac{20}{\sqrt{100}}$
$\text{Distance}=\dfrac{20}{10}=2\text{ units}$
SOLUTION 2
The distance from a point $(h,k)$ to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substitute the origin $(0,0)$:
$\text{Distance}=\dfrac{|-8(0)-6(0)+20|}{\sqrt{(-8)^2+(-6)^2}}$
$\text{Distance}=\dfrac{20}{\sqrt{100}}=2\text{ units}$
3. Find the distance from the origin to the line $-4x-3y=40$.
ANSWER: $8$ units
SOLUTION 1
Rule: the distance from the origin to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=-4$, $b=-3$, $c=-40$ (writing the line as $ax+by+c=0$):
$\text{Distance}=\dfrac{|-40|}{\sqrt{(-4)^2+(-3)^2}}$
$\text{Distance}=\dfrac{40}{\sqrt{25}}$
$\text{Distance}=\dfrac{40}{5}=8\text{ units}$
SOLUTION 2
The distance from a point $(h,k)$ to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substitute the origin $(0,0)$:
$\text{Distance}=\dfrac{|-4(0)-3(0)-40|}{\sqrt{(-4)^2+(-3)^2}}$
$\text{Distance}=\dfrac{40}{\sqrt{25}}=8\text{ units}$
4. Find the distance from the origin to the line $5x+12y=39$.
ANSWER: $3$ units
SOLUTION 1
Rule: the distance from the origin to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=5$, $b=12$, $c=-39$ (writing the line as $ax+by+c=0$):
$\text{Distance}=\dfrac{|-39|}{\sqrt{5^2+12^2}}$
$\text{Distance}=\dfrac{39}{\sqrt{169}}$
$\text{Distance}=\dfrac{39}{13}=3\text{ units}$
SOLUTION 2
The distance from a point $(h,k)$ to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substitute the origin $(0,0)$:
$\text{Distance}=\dfrac{|5(0)+12(0)-39|}{\sqrt{5^2+12^2}}$
$\text{Distance}=\dfrac{39}{\sqrt{169}}=3\text{ units}$
5. Find the distance from the origin to the line $-12x-9y=15$.
ANSWER: $1$ units
SOLUTION 1
Rule: the distance from the origin to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=-12$, $b=-9$, $c=-15$ (writing the line as $ax+by+c=0$):
$\text{Distance}=\dfrac{|-15|}{\sqrt{(-12)^2+(-9)^2}}$
$\text{Distance}=\dfrac{15}{\sqrt{225}}$
$\text{Distance}=\dfrac{15}{15}=1\text{ units}$
SOLUTION 2
The distance from a point $(h,k)$ to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substitute the origin $(0,0)$:
$\text{Distance}=\dfrac{|-12(0)-9(0)-15|}{\sqrt{(-12)^2+(-9)^2}}$
$\text{Distance}=\dfrac{15}{\sqrt{225}}=1\text{ units}$
6. Find the distance from the origin to the line $-9x+12y=15$.
ANSWER: $1$ units
SOLUTION 1
Rule: the distance from the origin to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|c|}{\sqrt{a^2+b^2}}$
Given $a=-9$, $b=12$, $c=-15$ (writing the line as $ax+by+c=0$):
$\text{Distance}=\dfrac{|-15|}{\sqrt{(-9)^2+12^2}}$
$\text{Distance}=\dfrac{15}{\sqrt{225}}$
$\text{Distance}=\dfrac{15}{15}=1\text{ units}$
SOLUTION 2
The distance from a point $(h,k)$ to $ax+by+c=0$ is:
$\text{Distance}=\dfrac{|ah+bk+c|}{\sqrt{a^2+b^2}}$
Substitute the origin $(0,0)$:
$\text{Distance}=\dfrac{|-9(0)+12(0)-15|}{\sqrt{(-9)^2+12^2}}$
$\text{Distance}=\dfrac{15}{\sqrt{225}}=1\text{ units}$