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Coordinate Geometry: Practice Worksheet

Distance, section and midpoint formulae, coordinate relationships, and geometric reasoning.

18 problems·35–45 min·★★★★★
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  1. Problem 1
    Section Formula·★★★★☆
    Point $P(a,b)$ divides the segment joining $A(3, 4)$ and $B(11, -4)$ internally in the ratio $4:4$. Find $a$ and $b$.
    ▶Answer
    $a=7$ and $b=0$
    ▶Step-by-step solution
    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 $7$.
    3. The y-coordinate simplifies to $0$, so $P=(7, 0)$.
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  2. Problem 2
    Section Formula·★★★★☆
    Find the ratio in which $P(5, 5)$ divides the segment joining $A(1, 2)$ and $B(9, 8)$.
    ▶Answer
    $AP:PB=1:1$
    ▶Step-by-step solution
    1. Compare the coordinate displacement from A to P with that from P to B.
    2. $\overrightarrow{AP}=1$ copies of the basic direction vector, while $\overrightarrow{PB}=1$ copies.
    3. Hence $AP:PB=1:1$.
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  3. Problem 3
    Distance Formula·★★☆☆☆
    Find the distance of $P(7, 6)$ from the y-axis.
    1. (A)$6$ units
    2. (B)$13$ units
    3. (C)$7$ units
    4. (D)$1$ units
    ▶Answer
    Option C: $7$ units
    ▶Step-by-step solution
    1. Distance from the y-axis is the absolute value of the x-coordinate.
    2. That coordinate is $7$.
    3. Hence the distance is $7$ units.
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  4. Problem 4
    Distance Formula·★★☆☆☆
    Find the distance between $A(-3, -5)$ and $B(-3, -1)$.
    1. (A)$4$ units
    2. (B)$1$ units
    3. (C)$6$ units
    4. (D)$5$ units
    ▶Answer
    Option A: $4$ units
    ▶Step-by-step solution
    1. The points have the same x-coordinate, so the segment is vertical.
    2. Its length is $|-1-(-5)|$.
    3. Therefore the distance is $4$ units.
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  5. Problem 5
    Distance Formula·★★☆☆☆
    Find the distance of $P(5, -12)$ from the origin.
    1. (A)$5$ units
    2. (B)$17$ units
    3. (C)$13$ units
    4. (D)$12$ units
    ▶Answer
    Option C: $13$ units
    ▶Step-by-step solution
    1. Use $OP=\sqrt{x^2+y^2}$.
    2. $OP=\sqrt{(5)^2+(-12)^2}=\sqrt{169}$.
    3. Thus $OP=13$ units.
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  6. Problem 6
    Distance Formula·★★★☆☆
    Find the distance between $(-5, 3)$ and $(-2, 4)$.
    1. (A)$\sqrt{8}$ units
    2. (B)$\sqrt{10}$ units
    3. (C)$3$ units
    4. (D)$4$ units
    ▶Answer
    Option B: $\sqrt{10}$ units
    ▶Step-by-step solution
    1. Use the distance formula.
    2. The squared distance is $(-2--5)^2+(4-3)^2=10$.
    3. Hence the distance is $\sqrt{10}$ units.
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  7. Problem 7
    Distance Formula·★★★☆☆
    Rectangle AOBC has $O(0,0)$, $A(0,6)$ and $B(7,0)$. Find the length of its diagonal.
    1. (A)$\sqrt{85}$ units
    2. (B)$7$ units
    3. (C)$6$ units
    4. (D)$13$ units
    ▶Answer
    Option A: $\sqrt{85}$ units
    ▶Step-by-step solution
    1. The horizontal and vertical side lengths are 7 and 6 units.
    2. By Pythagoras, the squared diagonal is $7^2+6^2=85$.
    3. Therefore the diagonal is $\sqrt{85}$ units.
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  8. Problem 8
    Distance Formula·★★★☆☆
    Find the perimeter of the triangle with vertices $(0,0)$, $(3,0)$ and $(0,4)$.
    1. (A)$13$ units
    2. (B)$7$ units
    3. (C)$12$ units
    4. (D)$5$ units
    ▶Answer
    Option C: $12$ units
    ▶Step-by-step solution
    1. The perpendicular sides have lengths 3 and 4 units.
    2. The third side is $\sqrt{3^2+4^2}=5$ units.
    3. The perimeter is $3+4+5=12$ units.
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  9. Problem 9
    Distance Formula·★★★☆☆
    The distance between $A(5, -1)$ and $B(13,p)$ is $17$ units. Find $p$.
    1. (A)$14$ only
    2. (B)$-16$ or $14$
    3. (C)$-1$
    4. (D)$-16$ only
    ▶Answer
    Option B: $-16$ or $14$
    ▶Step-by-step solution
    1. $(13-5)^2+(p--1)^2=17^2$.
    2. Thus $(p--1)^2=225=225$.
    3. So $p=-16$ or $p=14$.
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  10. Problem 10
    Distance Formula·★★★★☆
    Find the points on the x-axis that are $\sqrt{58}$ units from $A(-4, 3)$. How many are there?
    ▶Answer
    The points are $(-11, 0)$ and $(3, 0)$; there are two.
    ▶Step-by-step solution
    1. Let the point be $(x,0)$. Then $(x--4)^2+(0-3)^2=58$.
    2. So $(x--4)^2=49$ and $x=-11$ or $x=3$.
    3. Both values give points on the x-axis, so there are two points.
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  11. Problem 11
    Distance Formula·★★★★☆
    The distance between $A(-4, 0)$ and $B(a,15)$ is $17$ units. Given that $a>-4$, find $a$.
    ▶Answer
    $a=4$
    ▶Step-by-step solution
    1. $(a--4)^2+(15-0)^2=17^2$.
    2. Thus $(a--4)^2=64=64$.
    3. Since $a>-4$, take the positive displacement: $a=-4+8=4$.
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  12. Problem 12
    Circles·★★★★☆
    A circle is centred at the origin with radius 9 units. Which point does not lie in its interior?
    1. (A)$(2, 3)$
    2. (B)$(9, 4)$
    3. (C)$(3, 1)$
    4. (D)$(1, 2)$
    ▶Answer
    Option B: $(9, 4)$
    ▶Step-by-step solution
    1. Interior points satisfy $x^2+y^2<81$.
    2. For $(9, 4)$, $x^2+y^2=97>81$.
    3. So that point is outside the circle.
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  13. Problem 13
    Circles·★★★★☆
    Write True or False and justify: A circle is centred at the origin with radius 4 units. The point $Q(1, 1)$ lies outside it.
    ▶Answer
    False.
    ▶Step-by-step solution
    1. Compare $OQ^2$ with $r^2=16$.
    2. Here $OQ^2=2$.
    3. Since $OQ^2<r^2$, Q is inside.
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  14. Problem 14
    Circles·★★★★★
    Write True or False and justify: $P(7, -2)$ lies on the circle with centre $C(2, -2)$ and radius 5 units.
    ▶Answer
    True.
    ▶Step-by-step solution
    1. Compute $CP^2=25$.
    2. The squared radius is $25$.
    3. They are equal, so P lies on the circle.
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  15. Problem 15
    Circles·★★★★★
    A circle has centre $(2a,a-7)$, passes through $P(-1, 2)$, and has radius 13 units. Find all possible values of a.
    ▶Answer
    $a=-3$ or $a=\frac{29}{5}$
    ▶Step-by-step solution
    1. The distance from the centre to P equals the radius.
    2. Substitute $(2a,a-7)$ and P into the squared-distance equation $CP^2=169$.
    3. Solving the resulting quadratic gives $a=-3$ or $a=\frac{29}{5}$.
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  16. Problem 16
    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
    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.
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  17. Problem 17
    Midpoint Formula·★★☆☆☆
    If $P(a/4, -\frac{5}{2})$ is the midpoint of $Q(0, -2)$ and $R(6, -3)$, find $a$.
    1. (A)$12$
    2. (B)$-12$
    3. (C)$8$
    4. (D)$16$
    ▶Answer
    Option A: $12$
    ▶Step-by-step solution
    1. The x-coordinate of the midpoint is $(0+6)/2=3$.
    2. Thus $a/4=3$.
    3. Therefore $a=12$.
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  18. Problem 18
    Section Formula·★★☆☆☆
    Point $P(-3, 1)$ lies on segment $AB$, where $A(-4, 0)$ and $B(-1, 3)$. Which relation is correct?
    1. (A)$PB=\frac{1}{3}AB$
    2. (B)$AP=\frac{1}{3}AB$
    3. (C)$AP=PB$
    4. (D)$AP=\frac{2}{3}AB$
    ▶Answer
    Option B: $AP=\frac{1}{3}AB$
    ▶Step-by-step solution
    1. From A to B the coordinate change is 3 copies of the basic direction vector.
    2. From A to P it is 1 copies of the same vector.
    3. Therefore $AP/AB=1/3$.
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