# Coordinate Geometry Problems - Page 2

Coordinate Geometry Problems
• Page 2
11.
Use parallelogram ABCD to check which one of the following is true.

 a. AB = AD b. AB = BC c. AB = DC d. AB = AC

#### Solution:

Use the distance formula, d = (x2 -x1)² + (y2 -y1)²

AB = (a - 0)² + (0 - 0)²  = a²  = a

BC = (a + b - a)² + (c - 0)²  = b² + c², so AB ≠ BC

AD = (b - 0)² + (c - 0)²  = b² + c² , so AB ≠ AD

DC = (a + b - b)² + (c - c)²  = a² + 0 = a, So AB = DC

AC = (a + b - 0)² + (c - 0)²  = (a + b)² + c²  , so AB ≠ AC.

So, AB = DC.

12.
Use parallelogram ABCD to check which one of the following is true.

 a. BC = AD b. BC = BD c. BC = DC d. BC = AB

#### Solution:

Use the distance formula, d = (x2 -x1)² + (y2 -y1)²

BC = (a + b - a)² + (c - 0)²  = b² + c²

AB = (a - 0)² + (0 - 0)²  = a² + 0²  = a, so BC ≠ AB

AD = (b - 0)² + (c - 0)²  = b² + c² , so BC = AD.

13.
Use parallelogram OPQR to check which one of the following is true.

 a. OQ = PR b. OP = PR c. OP ≠ RQ d. OQ ≠ PR

#### Solution:

Use the distance formula, d = (x2 -x1)² + (y2 -y1)²

OP = (x - 0)² + (0 - 0)²  = x² + 0²  = x

PR = (a - x)² + (b - 0)²  = (a - x)² + b² , so OP ≠ PR

RQ = (x + a - a)² + (b - b)²  = x² + 0²  = x, So OP = RQ.

OQ = (x + a - 0)² + (b - 0)²  = (x + a)² + b² , So, OQ ≠ PR

14.
(-9,0), (9,10) and (8,-10) are the vertices of ___________.
 a. A scalene triangle b. An equilateral triangle c. An isosceles triangle d. A right triangle

#### Solution:

Let (-9, 0), (9, 10) and (8, -10) be A, B and C respectively.

Use the distance formula, d = (x2-x1)²+(y2-y1)²

AB = (9-(- 9))2+(10 - 0)2 = 182+102 = 424 = 20.59

BC = (8 - 9)2+(- 10 - 10)2 = (-1)2+(-16)2 = 257 = 16.03

CA = (-9-8)2 +(0-(-10))2 = (-17)2+102 = 389 = 4.24

So, AB ≠ BC ≠ CA

The coordinates (-9,0), (9,10) and (8, -10) forms a scalene triangle.

15.
ABCD is a quadrilateral with vertices A(0, 0), B(-$a$, 0), C(-$a$, $a$) and D(0, $a$). The midpoint of $\stackrel{‾}{\mathrm{AC}}$ is same as the midpoint of _____________.
 a. $\stackrel{‾}{\mathrm{AD}}$ b. $\stackrel{‾}{\mathrm{BD}}$ c. $\stackrel{‾}{\mathrm{AB}}$ d. $\stackrel{‾}{\mathrm{BC}}$

#### Solution:

Use the midpoint formula (x1 +x22 , y1 +y22).

Midpoint of AC: (0 - a2 , 0 + a2) = (- a2 , a2)

Midpoint of BD: (- a + 02 , 0 + a2) = (- a2 , a2)

Midpoint of AB: (0 - a2 , 0 + 02) = (- a2 , 0)

Midpoint of BC: (- a - a2 , 0 + a2) = (- 2a2 , a2) = (- a , a2)

Midpoint of AD: (0 + 02 , 0 + a2) = (0 , a2)

So, midpoint of AC is same as the midpoint of BD.

16.
A (-2,5), B (3, -4) and C (7,10) are the vertices of a right triangle with BC as the hypotenuse. The length of median AD to the hypotenuse is equal to the length of ____________.
 a. $\frac{1}{3}$BC b. $\frac{1}{2}$ BC c. BC d. 2 BC

#### Solution:

A median of a triangle is a line segment from a vertex to the midpoint of the opposite side.

Coordinates of D: (3 + 72 , -4 + 102) = (5, 3)
[Use the mid point formula.]

Use the distance formula, d = (x2 -x1)² + (y2 -y1)² to find the lengths of AD and BC.

AD = [5 - (-2)]2 +(3 - 5)2 = 72 +(-2)2 = 53

BC = (7 - 3)2 +[10 - (-4)]2 = 42 +142 = 253

17.
The points A (2,4), B (2,6) and C (2 + $\sqrt{3}$, 5) are the vertices of an equilateral triangle ABC. If D, E and F are the midpoints of the sides AB, BC and CA respectively, then triangle DEF forms _____________.
 a. a scalene triangle b. an equilateral triangle c. a right triangle d. an obtuse triangle

#### Solution:

Use the midpoint formula (x1 +x22) , (y1 +y22) to find coordinates of D, E and F.

Coordinates of D: (2 + 22 , 4 + 62) = (2,5)
[Use the mid point formula.]

Coordinates of E: (2 + 2 +32 , 6 + 52) = (4 +32 , 112)
[Use the mid point formula.]

Coordinates of F: (2 + 2 +32 , 4 + 52) = (4 +32 , 92)
[Use the mid point formula.]

Use the distance formula, d = (x2 -x1)² + (y2 -y1)² to find the lengths of DE, EF and DF.

DE = (4 +32 - 2)2 + (112 - 5)2

= (32)2 + (12)2 = 34 +14 = 1

EF = (4 +32 -4 +32)2 + (92 -112)2

= 02 +(-1)2 = 1

DF = (4 +32 - 2)2 + (92 - 5)2

= (32)2 + (-12)2 = 34 +14 = 1

The lengths of DE, EF and DF are the same.

So, the triangle formed with the vertices D, E and F is an equilateral triangle.

18.
If ABCD is a Rhombus with vertices A(0,0), B(5,0), C(8,4) and D(3,4), then the sum of the squares of the lengths of the four sides is equal to which of the following?
 a. Sum of the squares of the lengths of the diagonals. b. Difference of the lengths of the diagonals. c. Difference of the squares of the lengths of the diagonals. d. Sum of the lengths of the diagonals.

#### Solution:

Use the distance formula, d = (x2 -x1)² + (y2 -y1)² to find the lengths of the sides and the diagonals.

AB = (5 - 0)2 +(0 - 0)2 = 52 +02 = 5

Since ABCD is a Rhombus, all the sides are equal in length.

So, the sum of the squares of lengths of the four sides is equal to 4·AB2 = 4(5)2 = 100.

AC = (8 - 0)2 +(4 - 0)2 = 82 +42 = 80

BD = (3 - 5)2 +(4 - 0)2 = (-2)2 +42 = 20

AC2 + BD2 = (802) + (202) = 80 + 20 = 100

So, the sum of the squares of the lengths of four sides of a Rhombus is equal to the sum of the squares of the lengths of the diagonals.

19.
Which one of the following can be the vertices of a square?
 a. (0, 0), (3, 0), (3, 3), (0, 3) b. (0, 0), (5, 0), (3, 3), (0, 3) c. (0, 0), (3, 0), (5, 4), (2, 4) d. (0, 0), (4, 0), (4, 3), (0, 3)

#### Solution:

Plot the points (0, 0), (4, 0), (4, 3) and (0, 3) in the coordinate plane.

These are the vertices of a rectangle as the opposite sides are of equal length and the interior angles are right angles.

Plot the points (0, 0), (3, 0), (5, 4) and (2, 4) in the coordinate plane.

These are the vertices of a parallelogram as the opposite sides and diagonally opposite angles are equal.

Plot the points (0, 0), (3, 0), (3, 3), (0, 3) in the coordinate plane

These are the vertices of a square as the lengths of all the sides are equal and the diagonals are of equal length.

Plot the points (0, 0), (5, 0), (3, 3) and (0, 3) in the coordinate plane.

These are the vertices of a trapezoid as one pair of opposite sides are parallel.

Another method:

Let (0, 0), (3, 0), (3, 3) and (0, 3) be the coordinates of the vertices A, B, C, and D.

We use the distance formula, d = (x2 -x1)² + (y2 -y1)² to fnd the lengths of AB, BC, CD and DA.

AB = (3 - 0)2 +(0 - 0)2 = 32 +02 = 3

BC = (3 - 3)2 +(0 - 3)2 = 02 +(-3)2 = 9 = 3

CD = (0 - 3)2 +(3 - 3)2 = (-3)2 +02 = 9 = 3

DA = (0 - 0)2 +(0 - 3)2 = 02 +(-3)2 = 9 = 3

The lengths of AB, BC, CD, and DA are equal.

AC = (3 - 0)2 +(3 - 0)2 = 18 = 32

BD = (0 - 3)2 +(3 - 0)2 = (-3)2 +32 = 18 = 32

The lengths of AC and BD are the same.

The lengths of all the sides are equal and diagonals are also equal. So, the points A, B, C, and D form a square.

20.
The end points of a diameter AB of a circle are (-1, -3) and (5, 5). P(-2, 4) is any other point on the the circle then the angle between $\stackrel{‾}{\mathrm{PA}}$ and $\stackrel{‾}{\mathrm{PB}}$ is _______.
 a. a right angle b. a reflex angle c. an obtuse angle d. an acute angle

#### Solution:

Use the formula : y2 -y1x2 -x1 to find slope of PA and PB

Slope PA = 4 - (-3)-2 - (-1) = 7-1 = - 7

Slope PB = 4 - 5-2 - 5 = 17

(Slope PA) x (slope PB) = -7 x 17= -1

As the slopes of two lines PA and PB are negative reciprocals of each other, the lines PA and PB are perpendicular.

So, angle between the lines PA and PB is a right angle.