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Neokaal Study

Coordinate Geometry

Competency Practice

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

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  1. 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 $D$.

  2. 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 $p$.

  3. Find a point on the $y$-axis which is equidistant from the points $A(6, 5)$ and $B(-4, 3)$.

  4. Point $P(a,b)$ divides the segment joining $A(5, 0)$ and $B(8, -3)$ internally in the ratio $1:2$. Find $a$ and $b$.

  5. Find the ratio in which $P(9, 10)$ divides the segment joining $A(-3, 1)$ and $B(17, 16)$.

  6. 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.

  7. Find the points on the x-axis that are $\sqrt{50}$ units from $A(7, 5)$. How many are there?

  8. 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

Neokaal Study

Answer Key

  1. (i) $13\text{ km}$, (ii) $M(6, 2.5)$, (iii) $5\text{ km}$

    1. (i) Straight-line flight distance $AB = \sqrt{(12 - 0)^2 + (5 - 0)^2} = \sqrt{169} = 13\text{ km}$.
    2. (ii) Recharging station coordinates $M = \left(\frac{0 + 12}{2}, \frac{0 + 5}{2}\right) = (6, 2.5)$.
    3. (iii) Distance $MD = \sqrt{(2 - 6)^2 + (5 - 2.5)^2} = \sqrt{25} = 5\text{ km}$.

  2. Ratio $= 1 : 2$, $p = -\frac{5}{3}$

    1. Let $P(p, -2)$ divide segment $AB$ in the ratio $k : 1$. Using the Section Formula: $y = \frac{k(2) + 1(-4)}{k + 1} = -2$.
    2. Solving gives $(2 - -2)k = -2 - (-4) \implies k = \frac{1}{2}$. Thus $P$ divides $AB$ in the ratio $1 : 2$.
    3. Using this ratio for the $x$-coordinate: $p = \frac{1(1) + 2(-3)}{1 + 2} = -\frac{5}{3}$.

  3. $(0, 9)$

    1. Any point on the $y$-axis has coordinates $P(0, y)$.
    2. Equating distances $PA^2 = PB^2$ gives $(0 - 6)^2 + (y - 5)^2 = (0 - (-4))^2 + (y - 3)^2$.
    3. Expanding gives $y^2 - 10y + 61 = y^2 - 6y + 25$.
    4. Simplifying gives $-4y = -36 \implies y = 9$.
    5. Therefore, the required equidistant point is $(0, 9)$.

  4. $a=6$ and $b=-1$

    1. Use $P=((n x_1+m x_2)/(m+n),(n y_1+m y_2)/(m+n))$.
    2. The x-coordinate simplifies to $6$.
    3. The y-coordinate simplifies to $-1$, so $P=(6, -1)$.

  5. $AP:PB=3:2$

    1. Compare the coordinate displacement from A to P with that from P to B.
    2. $\overrightarrow{AP}=3$ copies of the basic direction vector, while $\overrightarrow{PB}=2$ copies.
    3. Hence $AP:PB=3:2$.

  6. $y=-4$ or $y=4$; the corresponding PQ distances are $10$ and $6$ units.

    1. Set $AP^2=AQ^2$.
    2. This gives $y^2=16$, so $y=-4$ or $y=4$.
    3. Use the distance formula between P and Q for each value.

  7. The points are $(2, 0)$ and $(12, 0)$; there are two.

    1. Let the point be $(x,0)$. Then $(x-7)^2+(0-5)^2=50$.
    2. So $(x-7)^2=25$ and $x=2$ or $x=12$.
    3. Both values give points on the x-axis, so there are two points.

  8. Option B: $2\sqrt{5}$ units

    1. Use the distance formula.
    2. The squared distance is $(4-2)^2+(4-0)^2=20$.
    3. Hence the distance is $2\sqrt{5}$ units.