﻿ Coordinate Geometry Problems | Problems & Solutions Coordinate Geometry Problems

Coordinate Geometry Problems
• Page 1
1.
Find the perimeter of quadrilateral ABCD.  a. 24.08 units b. 28 units c. 30.25 units d. 30 units

Solution:

Distance between two points (x1, y1) and (x2, y2) = (x2-x1)2+(y2 -y1)2
[Formula.]

AB = (11-3)²+(8-8)²
[Replace (x1, y1) with (3, 8) and (x2, y2) with (11, 8).]

AB = 82+0 = 8
[Simplify.]

BC = (8-11)²+(2-8)²
[Substitute the values.]

BC = (-3)²+(-6)²=35
[Simplify.]

CD = (3-8)²+(4-2)²
[Substitute the values.]

CD = (-5)²+2²=29
[Simplify.]

DA = (3-3)²+(8-4)²
[Substitute the values.]

DA = 0²+4² = 4
[Simplify.]

Perimeter of ABCD = AB + BC + CD + DA

Perimeter = 8 + 35+29 + 4
[Substitute the values of AB, BC, CD, DA and add.]

= 8 + 6.69 + 5.39 + 4 = 24.08

The perimeter of ABCD = 24.08 units.

2.
(-2, 5), (3, - 4) and (7,10) are the vertices of _____________. a. An isosceles triangle b. A right triangle c. An equilateral triangle d. A scalene triangle

Solution:

Let (-2, 5), (3, - 4) and (7, 10) be A, B and C respectively.

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

AB = (3 - (- 2))2 +(-4 - 5)2 = 52 +(-9)2 = 106

BC = (7 - 3)2 +(10 - (-4))2 = 42 +142 = 212

CA = (-2 - 7)2 +(5 - 10)2 = (-9)2 +(-5)2 = 106

AB2 + CA2 =106 + 106 = 212 and BC2 = 212, So, BC2 = AB2 + CA2

Therefore, the vertices (- 2, 5), (3, - 4) and (7, 10) form a right triangle.

3.
If A(0,0), B(5, 0), C(8, 4) and D(3, 4) are the vertices of a rhombus, then angle between the diagonals of it is _________. a. a reflex angle b. an acute angle c. an obtuse angle d. a right angle

Solution:

Use the formula: y2 -y1x2 -x1 to find the slope of AC and BD.

Slope AC = 4 - 08 - 0 = 48 = 12

Slope BD = = 4 - 03 - 5 = 4-2 = - 2

(SlopeAC) × (Slope BD) = 12 × -2 = -1

As the slopes of the two line segments AC and BD are negative reciprocals of each other, the line segments AC and BD are perpendicular.

So the angle between the diagonals of a rhombus is a right angle.

4.
(7, 7), (6, 4) and (4, 6) are the vertices of ________________. a. an equilateral triangle b. an isosceles triangle c. a scalene triangle d. a right triangle

Solution:

Let (7, 7), (6, 4) and (4, 6) be A, B and C respectively.

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

AB = (6 - 7)2 +(4 - 7)2 = (-1)2 +(-3)2 = 10

BC = (4 - 6)2 +(6 - 4)2 = (-2)2 +22 = 8

CA = (7 - 4)2 +(7 - 6)2 = 32 +12 = 10

The length of AB and CA are equal.

So, the vertices (6, 6), (5, 2) and (2, 5) form an isosceles triangle.

5.
(0, 0), (4, 0) and (2, 2$\sqrt{3}$) are the vertices of ___________. a. an equilateral triangle b. an isosceles triangle c. a right triangle d. a scalene triangle

Solution:

Let (0, 0), (4, 0) and (2, 23) be P, Q and R respectively.

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

PQ = (4 - 0)2 +(0 - 0)2 = 42 +02 = 4

QR = (2 - 4)2 +(23 - 0)2 = (-2)2 +(23)2 = 16 = 4

RP = (0 - 2)2 +(0 - 23)2 = (-2)2 +(-23)2 = 4 + 12 = 4

The lengths of PQ, QR and RP are equal.

So, the vertices (0, 0), (4, 0) and (2, 23) form an equilateral triangle.

6.
The line segment joining the midpoints of two sides of any triangle is _________. a. intersects the third side. b. parallel to the third side c. perpendicular to the third side.

Solution:

Let (0, 0), (x, 0) and (a, b) be the coordinates of vertices A,B and C of triangle ABC respectively.

Let D and E be the midpoints of AC and BC, respectively.

Coordinates of D: (0 + a2 , 0 + b2) = (a2 , b2)
[Use midpoint formula.]

Coordinates of E: (x + a2 , 0 + b2) = (x + a2 , b2)

Slope DE = b2 -b2x + a2 -a2 = 0x2 = 0

Slope AB = 0 - 0x - 0 = 0x = 0

Slope DE and AB are equal.

The line segment joining D and E is parallel to the side AB.
[Two lines are parallel if they have equal slopes.]

So, the line segment joining the midpoints of two sides of any triangle is parallel to the third side.

7.
A quadrilateral with vertices (0, 0), (10, 0), (16, 8) and (6, 8) forms a ________. a. Square b. Rectangle c. Rhombus d. Parallelogram

Solution:

Let (0, 0), (10, 0), (16, 8) and (6, 8) be A, B, C and D respectively.

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

AB = (10 - 0)2 +(0 - 0)2 = 10

BC = (16 - 10)2 +(8 - 0)2 = 62 +82 = 100 = 10

CD = (6 - 16)2 +(8 - 8)2 = (-10)2 +02 = 100 = 10

DA = (0 - 6)2 +(0 - 8)2 = (-6)2 +(-8)2 = 100 = 10

AC = (16 - 0)2 +(8 - 0)2 = 162 +82 = 320

BD = (6 - 10)2 +(8 - 0)2 = (-4)2 +82 = 32

The lengths of the sides AB, BC, CD and DA are equal and the lengths of the diagonals AC and BD are not equal in quadrilateral ABCD.

So, ABCD is a rhombus.

8.
A quadrilateral PQRS with vertices (2,-2), (8,4),(5,7) and (-1,1) forms a ______________. a. Rhombus b. Square c. Rectangle

Solution:

Use the distance formula, d = (x2 -x1)² + (y2 -y1)² to find the lengths of PQ, QR, RS, PS, PR and QS.

PQ = (8 - 2)2 +(4 - (-2))2
[(x1, y1) = (2, -2); (x2, y2) = (8, 4).]

= 62 +62 = 72 =62

QR = (5 - 8)2 +(7 - 4)2
[(x1, y1) = (8, 4); (x2, y2) = (5, 7).]

= (-3)2 +32 = 18 = 32

PR = (5 - 2)2 +(7 - (-2))2
[( x1, y1) = (2, -2); (x2, y2) = (5, 7).]

= 32 +92 = 90 = 310

RS = (-1 - 5)2 +(1 - 7)2
[(x1, y1) = (5, 7); (x2, y2) = (-1, 1).]

= (-6)2 +(-6)2 = 72 = 62

PS = (-1 - 2)2 +(1 - (-2))2
[(x1, y1) = (2, -2); (x2, y2) = (-1, 1).]

= (-3)2 +32 = 18 = 32

QS = (-1 - 8)2 +(1 - 4)2
[( x1, y1) = (8, 4); (x2, y2) = (-1, 1).]

= (-9)2 +(-3)2 = 90 = 310

PQ = RS, QR = PS and PR = QS. Therefore, PQRS forms a Rectangle.

9.
A (- 4, 3), B (0, 9), C (-3, 11) are the three consecutive vertices of a parallelogram ABCD. Which one of the following can be the coordinates of fourth vertex D? a. (5, -7) b. (7, -5) c. (-7, 5) d. (-5, 7)

Solution:

ABCD is a parallelogram.

So, the diagonals AC and BD bisect each other.

Let (a,b) be the coordinates of D.

Midpoint of AC = Midpoint of BD

(-4 + (-3)2 , 3 + 112) = (0 + a2 , 9 + b2)

(-72 , 7) = (a2 , 9 + b2)

-72 = a2 and 7 = 9 + b2

-7 = a and 5 = b

So, the coordinates of D are (-7, 5).

10.
The base end points of an isosceles triangle are (-2, 4) and (6, 0). Which of the following can not be the coordinates of the vertex of the triangle? a. (5, 8) b. (-1, - 4) c. (1, 0) d. (0, 1)

Solution:

Let (-2, 4) and (6, 0) be the coordinates of A and B.

Slope AB: 0 - 46 - (-2) = -48 = -12

Midpoint M: (-2 + 62 , 4 + 02) = (2, 2)

The median to the base of an isosceles triangle is perpendicular to the base.

So, find an equation of the line that is perpendicular to AB and passes through its midpoint M.

y - y1 = m(x - x1)
[(x1, y1) are the coordinates of M.]

y - 2 = 2 (x - 2)
[m is the negative reciprocal of -12, the slope of AB.]

y = 2x - 2

- 4 = 2(-1) - 2
[Substitute the values.]

- 4 = - 4 is a true statement.

0 = 2(1) - 2
[Substitute the values.]

0 = 0 is a true statement

8 = 2(5) - 2
[Substitute the values.]

8 = 8 is a true statement.

1 = 2(0) - 2
[Substitute the values.]

1 = - 2 is false.

So, (0, 1) cannot be the coordinates of the vertex of an isosceles triangle whose base end points are (-2, 4) and (6, 0).