﻿ Coordinate Geometry Worksheet | Problems & Solutions Coordinate Geometry Worksheet

Coordinate Geometry Worksheet
• Page 1
1.
What is located at (3, 3)?  a. Palm tree b. Book store c. Food mart d. Pizza corner

Solution:

Start at zero.

As the first number in the ordered pair represents the number of spaces you have to move right, move 3 spaces to the right.

As the second number in the ordered pair represents the number of spaces you have to move up, move 3 spaces up.

So, the Food mart is located at the point (3, 3).

2.
ABCD is a square. The coordinates of C are (- 4, - 4). What are the coordiantes of A?  a. (- 8, - 8) b. (- 4, 4) c. (4, 4) d. (4, - 4)

Solution:

O is the origin.
[From the figure.]

ABCD is symmetrical with respect to x-axis and y-axis.
[ABCD is a square.]

D is the reflection of C w.r.t x-axis.
B is the reflection of C w.r.t y-axis.
A is the reflection of B w.r.t x-axis.

The coordinates are D(- 4, 4) , B(4, - 4) and A(4, 4).

The coordinates of A are A(4, 4).

3.
PQRS is a rectangle. The coordinates of M are M(8, 0) and N are N(0, 4). What are the coordinates of P and Q?  a. (8, 4 ) and (8, - 4 ) b. ( 8, - 4 ) and (- 8, 4 ) c. (-8, - 4 ) and (8, - 4) d. (- 8, - 4 ) and (- 8, 4 )

Solution:

PQRS is symmetrical w.r.t the coordinate axis.
[From the figure.]

Coordinates of R are R(8, 4).
[OM = 8, ON = 4.]

Coordinates of Q, P and S are Q(8, - 4), P(- 8, - 4) and S(- 8, 4).
[From Step 1.]

The Coordinates of P and Q are P(- 8, - 4) and Q(8, - 4).

4.
OPQR is a Rhombus. Find the coordinates of Q and R, if the coordinates of P are P(3, 6).  a. (- 6, 0) and ( 3 , - 6 ) b. (6, 0) and (- 3 , 6) c. ( 0 , 6) and ( - 3 , - 6 ) d. (6, 0) and (3 , - 6)

Solution:

Lines of symmetry for the given Rhombus are PR and OQ.

Coordinates of R are R(3, - 6)
[R is the reflection of P on x - axis.]

The coordinates of M are M(3, 0)
[M lies on x - axis.]

The coordinates of Q and R are Q(6, 0) and R(3, - 6)
[OM = QM.]

5.
ABCD is a parallelogram. The coordinates of B and C are B(3, - 3) and C(4 , 3). Find the coordinates of A.  a. ( 4, 3) b. (- 4, - 3) c. (- 4, 3) d. ( 4, - 3)

Solution:

Coordinates of C are (4, 3).
[Given data.]

O is the midpoint of AC.
[Diagonals of the parallelogram bisect each other.]

Coordinates of A are A(- 4, - 3).
[Coordinates of O are O(0,0). Apply midpoint formula.]

6.
OQ = 10. $m$$\angle$ORQ = 90, PM = 3 RM. Write the coordinates of P.  a. P(15, 5) b. P(5, 15) c. P(15, 15) d. P(5, 5)

Solution:

Coordinates of Q are Q(10, 0)
[OQ = 10.]

Coordinates of M are M(5 , 0)
[Longer diagonal of the kite is the perpendicular bisector.]

RM = QM.
[mORQ = 90°
mMRQ = 45°
ΔRMQ is isosceles right angle triangle.]

RM = 5 .

Coordinates of R are (5 , 5 )
[Step 2 & 4.]

PM = 3 RM
= 15.

Coordinates of P are P(5, 15 )
[Step 6 & step 2.]

7.
PQRS ia a parallelogram. Coordinates of Q and R are Q(8, 2) and R(4, - 2). What would be the coordinates of P, Q, R and S if the parallelogram is shifted so that SR is placed on the $x$-axis with S as the origin?  a. (- 4, 4), (- 16, - 4), (- 12, 0) and (0, 0) b. ( 4, - 4), (8 , 2 ) , (4, - 2 ) and (0, 0) c. (4, 4), (16, 4), (12, 0) and (0, 0) d. (- 4, 2 ), (16, 4) , (12, 0) and (- 8, - 2)

Solution:

The original position of P and S are P(- 4, 2) and S(- 8, - 2)
[Diagonals of the parallelogram bisect each other.]

When S coincides with the origin and SR is placed on x-axis, The coordinates of S and R are S(0,0) and R(12, 0)
[SR = 8 + 4 = 12.]

The new coordinates of P and Q are P(4, 4) and Q(16, 4)
[x-coordinates change to x+8, y-coordinates change to y+2.]

The new coordinates of P, Q , R and S are (4, 4) , (16, 4) , (12, 0) and (0, 0).

8.
The vertices of quadrilateral OABC are O(0, 0), A(28, 0), B(24, 8) and C(8, 24). Find the midpoint of the line joining the midpoints of $\stackrel{‾}{\mathrm{OA}}$ and $\stackrel{‾}{\mathrm{BC}}$.  a. (15, 8) b. (14, 0) c. (16, 16) d. (20, 24)

Solution: Let P, Q, R and S be midpoints of OA, AB, BC and OC.

The coordinates of P, Q, R and S are P(28 / 2, 0), Q(28+24 / 2, 8 / 2), R(24+8 / 2, 8+24 / 2) and S(8 / 2, 24 / 2).
[Midpoint of the line joining points (x1 , y1) and (x2 , y2) is (x2+x12, y2+y12).]

The coordinates of P and R are (14, 0) and (16, 16).

Midpoint of the line joining the midpoint of OA and BC is (15, 8).
[ P and R are the midpoints of OA and BC.]

9.
PQRS is a parallelogram. OP : OS = 1 : 2. Coordinates of P and Q are P(3, 0) and Q(9, 0). Find the coordinates of R.  a. (7, 6) b. (9, 6) c. (7, 7) d. (6, 6)

Solution:

OS = 21 × 3
[OP : OS = 1 : 2.]

Coordinates of S are S(0, 21× 3) = (0, 6 )
[S lies on y-axis.]

PQ = 9 - 3 = 6.
RS = 9 - 3 = 6.

x-coordiante of R = 6
[x-coordinate of S + PQ.]

y-coordinate of R = 6
[Same as y-coordinate of S.]

The coordinates of R are R(6, 6 )

10.
ABCD is an isosceles trapezoid. The coordinates of A and D are A (- 22, 0) and D (- 16, 10). Find the length of the line segment joining the midpoints of $\stackrel{‾}{\mathrm{AD}}$ and $\stackrel{‾}{\mathrm{BC}}$.  a. 19 b. 5 c. 38 d. 10

Solution:

y-axis is the line of symmetry of the trapezoid.

Coordinates of B and C are B(22, 0) and C(16, 10).
[B is the reflection of A w.r.t x-axis and C is reflection of D w.r.t y-axis.]

Midpoint of AD is (- 22-162 , 102 ) = (- 19, 5)
[Midpoint of line joining (x1, y1 ) and (x2 , y2) is (x2+x12, y2+y12).]

Midpoint of BC is (22+162 , 102 ) = (19, 5)
[Same as step 3.]

Length of the line segment joining the midpoint of AD and BC = (19 - (-19))2 = (38)2 = 38