Neokaal Study
Coordinate Geometry
Practice Worksheet
Distance, section and midpoint formulae, coordinate relationships, and geometric reasoning.
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Answer all questions.
Show all necessary working.
Use a separate notebook for your solutions.
Point $P(a,b)$ divides the segment joining $A(3, 4)$ and $B(11, -4)$ internally in the ratio $4:4$. Find $a$ and $b$.
Find the ratio in which $P(5, 5)$ divides the segment joining $A(1, 2)$ and $B(9, 8)$.
Find the distance of $P(7, 6)$ from the y-axis.
- (A)$6$ units
- (B)$13$ units
- (C)$7$ units
- (D)$1$ units
Find the distance between $A(-3, -5)$ and $B(-3, -1)$.
- (A)$4$ units
- (B)$1$ units
- (C)$6$ units
- (D)$5$ units
Find the distance of $P(5, -12)$ from the origin.
- (A)$5$ units
- (B)$17$ units
- (C)$13$ units
- (D)$12$ units
Find the distance between $(-5, 3)$ and $(-2, 4)$.
- (A)$\sqrt{8}$ units
- (B)$\sqrt{10}$ units
- (C)$3$ units
- (D)$4$ units
Rectangle AOBC has $O(0,0)$, $A(0,6)$ and $B(7,0)$. Find the length of its diagonal.
- (A)$\sqrt{85}$ units
- (B)$7$ units
- (C)$6$ units
- (D)$13$ units
Find the perimeter of the triangle with vertices $(0,0)$, $(3,0)$ and $(0,4)$.
- (A)$13$ units
- (B)$7$ units
- (C)$12$ units
- (D)$5$ units
The distance between $A(5, -1)$ and $B(13,p)$ is $17$ units. Find $p$.
- (A)$14$ only
- (B)$-16$ or $14$
- (C)$-1$
- (D)$-16$ only
Find the points on the x-axis that are $\sqrt{58}$ units from $A(-4, 3)$. How many are there?
The distance between $A(-4, 0)$ and $B(a,15)$ is $17$ units. Given that $a>-4$, find $a$.
A circle is centred at the origin with radius 9 units. Which point does not lie in its interior?
- (A)$(2, 3)$
- (B)$(9, 4)$
- (C)$(3, 1)$
- (D)$(1, 2)$
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.
Write True or False and justify: $P(7, -2)$ lies on the circle with centre $C(2, -2)$ and radius 5 units.
A circle has centre $(2a,a-7)$, passes through $P(-1, 2)$, and has radius 13 units. Find all possible values of a.
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.
If $P(a/4, -\frac{5}{2})$ is the midpoint of $Q(0, -2)$ and $R(6, -3)$, find $a$.
- (A)$12$
- (B)$-12$
- (C)$8$
- (D)$16$
Point $P(-3, 1)$ lies on segment $AB$, where $A(-4, 0)$ and $B(-1, 3)$. Which relation is correct?
- (A)$PB=\frac{1}{3}AB$
- (B)$AP=\frac{1}{3}AB$
- (C)$AP=PB$
- (D)$AP=\frac{2}{3}AB$
Neokaal Study
Answer Key
$a=7$ and $b=0$
- Use $P=((n x_1+m x_2)/(m+n),(n y_1+m y_2)/(m+n))$.
- The x-coordinate simplifies to $7$.
- The y-coordinate simplifies to $0$, so $P=(7, 0)$.
$AP:PB=1:1$
- Compare the coordinate displacement from A to P with that from P to B.
- $\overrightarrow{AP}=1$ copies of the basic direction vector, while $\overrightarrow{PB}=1$ copies.
- Hence $AP:PB=1:1$.
Option C: $7$ units
- Distance from the y-axis is the absolute value of the x-coordinate.
- That coordinate is $7$.
- Hence the distance is $7$ units.
Option A: $4$ units
- The points have the same x-coordinate, so the segment is vertical.
- Its length is $|-1-(-5)|$.
- Therefore the distance is $4$ units.
Option C: $13$ units
- Use $OP=\sqrt{x^2+y^2}$.
- $OP=\sqrt{(5)^2+(-12)^2}=\sqrt{169}$.
- Thus $OP=13$ units.
Option B: $\sqrt{10}$ units
- Use the distance formula.
- The squared distance is $(-2--5)^2+(4-3)^2=10$.
- Hence the distance is $\sqrt{10}$ units.
Option A: $\sqrt{85}$ units
- The horizontal and vertical side lengths are 7 and 6 units.
- By Pythagoras, the squared diagonal is $7^2+6^2=85$.
- Therefore the diagonal is $\sqrt{85}$ units.
Option C: $12$ units
- The perpendicular sides have lengths 3 and 4 units.
- The third side is $\sqrt{3^2+4^2}=5$ units.
- The perimeter is $3+4+5=12$ units.
Option B: $-16$ or $14$
- $(13-5)^2+(p--1)^2=17^2$.
- Thus $(p--1)^2=225=225$.
- So $p=-16$ or $p=14$.
The points are $(-11, 0)$ and $(3, 0)$; there are two.
- Let the point be $(x,0)$. Then $(x--4)^2+(0-3)^2=58$.
- So $(x--4)^2=49$ and $x=-11$ or $x=3$.
- Both values give points on the x-axis, so there are two points.
$a=4$
- $(a--4)^2+(15-0)^2=17^2$.
- Thus $(a--4)^2=64=64$.
- Since $a>-4$, take the positive displacement: $a=-4+8=4$.
Option B: $(9, 4)$
- Interior points satisfy $x^2+y^2<81$.
- For $(9, 4)$, $x^2+y^2=97>81$.
- So that point is outside the circle.
False.
- Compare $OQ^2$ with $r^2=16$.
- Here $OQ^2=2$.
- Since $OQ^2<r^2$, Q is inside.
True.
- Compute $CP^2=25$.
- The squared radius is $25$.
- They are equal, so P lies on the circle.
$a=-3$ or $a=\frac{29}{5}$
- The distance from the centre to P equals the radius.
- Substitute $(2a,a-7)$ and P into the squared-distance equation $CP^2=169$.
- Solving the resulting quadratic gives $a=-3$ or $a=\frac{29}{5}$.
$y=-4$ or $y=4$; the corresponding PQ distances are $10$ and $6$ units.
- 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.
Option A: $12$
- The x-coordinate of the midpoint is $(0+6)/2=3$.
- Thus $a/4=3$.
- Therefore $a=12$.
Option B: $AP=\frac{1}{3}AB$
- From A to B the coordinate change is 3 copies of the basic direction vector.
- From A to P it is 1 copies of the same vector.
- Therefore $AP/AB=1/3$.