Coordinate Geometry: Competency Practice

Drone navigation telemetry, internal division ratios, equidistant axis loci, and section parameters.

8 problems·20–25 min·★★★★☆

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Problem 1
coordinate-geometry·★★★★★
A delivery drone operates over a smart city grid where control hub $A$ is at $(0, 0)$ and destination hub $B$ is at $(12, 5)$, with all coordinates in kilometres. (i) Find the straight-line flight distance $AB$. (ii) Find the coordinates of the midpoint recharging station $M$. (iii) A sensor tower is located at $D(2, 5)$. Calculate the direct distance between $M$ and $D$.
▶Answer
(i) $13\text{ km}$, (ii) $M(6, 2.5)$, (iii) $5\text{ km}$
▶Step-by-step solution
(i) Straight-line flight distance $AB = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{169} = 13\text{ km}$.
(ii) Recharging station coordinates $M = \left(\frac{0 + 12}{2}, \frac{0 + 5}{2}\right) = (6, 2.5)$.
(iii) Distance $MD = \sqrt{(2 - 6)^2 + (5 - 2.5)^2} = \sqrt{25} = 5\text{ km}$.
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Problem 2
coordinate-geometry·★★★★☆
Find the ratio in which the point $P(p, -2)$ divides the line segment joining the points $A(-3, -4)$ and $B(1, 2)$. Hence, find the value of $p$.
▶Answer
Ratio $= 1 : 2$, $p = -\frac{5}{3}$
▶Step-by-step solution
Let $P(p, -2)$ divide segment $AB$ in the ratio $k : 1$. Using the Section Formula: $y = \frac{k(2) + 1(-4)}{k + 1} = -2$.
Solving gives $(2 - -2)k = -2 - (-4) \implies k = \frac{1}{2}$. Thus $P$ divides $AB$ in the ratio $1 : 2$.
Using this ratio for the $x$-coordinate: $p = \frac{1(1) + 2(-3)}{1 + 2} = -\frac{5}{3}$.
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Problem 3
coordinate-geometry·★★★★☆
Find a point on the $y$-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.
▶Answer
$(0, 9)$
▶Step-by-step solution
Any point on the $y$-axis has coordinates $P(0, y)$.
Equating distances $PA^2 = PB^2$ gives $(0 - 6)^2 + (y - 5)^2 = (0 - (-4))^2 + (y - 3)^2$.
Expanding gives $y^2 - 10y + 61 = y^2 - 6y + 25$.
Simplifying gives $-4y = -36 \implies y = 9$.
Therefore, the required equidistant point is $(0, 9)$.
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Problem 4
Section Formula·★★★★☆
Point $P(a,b)$ divides the segment joining $A(5, 0)$ and $B(8, -3)$ internally in the ratio $1:2$. Find $a$ and $b$.
▶Answer
$a=6$ and $b=-1$
▶Step-by-step solution
Use $P=((n x_1+m x_2)/(m+n),(n y_1+m y_2)/(m+n))$.
The x-coordinate simplifies to $6$.
The y-coordinate simplifies to $-1$, so $P=(6, -1)$.
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Problem 5
Section Formula·★★★★☆
Find the ratio in which $P(9, 10)$ divides the segment joining $A(-3, 1)$ and $B(17, 16)$.
▶Answer
$AP:PB=3:2$
▶Step-by-step solution
Compare the coordinate displacement from A to P with that from P to B.
$\overrightarrow{AP}=3$ copies of the basic direction vector, while $\overrightarrow{PB}=2$ copies.
Hence $AP:PB=3:2$.
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Problem 6
Equidistant Points·★★★★☆
Point $A(0, 0)$ is equidistant from $P(3, 4)$ and $Q(-3,y)$. Find the possible values of y and the corresponding distances PQ.
▶Answer
$y=-4$ or $y=4$; the corresponding PQ distances are $10$ and $6$ units.
▶Step-by-step solution
Set $AP^2=AQ^2$.
This gives $y^2=16$, so $y=-4$ or $y=4$.
Use the distance formula between P and Q for each value.
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Problem 7
Distance Formula·★★★☆☆
Find the points on the x-axis that are $\sqrt{50}$ units from $A(7, 5)$. How many are there?
▶Answer
The points are $(2, 0)$ and $(12, 0)$; there are two.
▶Step-by-step solution
Let the point be $(x,0)$. Then $(x-7)^2+(0-5)^2=50$.
So $(x-7)^2=25$ and $x=2$ or $x=12$.
Both values give points on the x-axis, so there are two points.
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Problem 8
Distance Formula·★★☆☆☆
Find the distance between $(2, 0)$ and $(4, 4)$.
1. (A)$4$ units
2. (B)$2\sqrt{5}$ units
3. (C)$\sqrt{12}$ units
4. (D)$6$ units
▶Answer
Option B: $2\sqrt{5}$ units
▶Step-by-step solution
Use the distance formula.
The squared distance is $(4-2)^2+(4-0)^2=20$.
Hence the distance is $2\sqrt{5}$ units.
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