# Coordinate Geometry Worksheet - Page 2

Coordinate Geometry Worksheet
• Page 2
11.
ABCD is a rectangle . The vertices of A are (- 8, - 3). Find the slope of the diagonal $\stackrel{‾}{\mathrm{AC}}$.

 a. $\frac{8}{3}$ b. c. $\frac{8}{9}$ d. $\frac{3}{8}$

#### Solution:

The coordinate axes are the lines of symmetry for the rectangle.

The coordinates of B, C and D are B(8, - 3), C(8, 3) and D(- 8, 3).

Slope of AC is = 3-(-3)8-(-8) = 616 = 38.
[Slope of line passing through (x1 , y1) and (x2 , y2) = y2-y1x2-x1.]

12.
OABC is a quadrilateral. The coordinates of A, B and C are A(28, 0), B(40, 20) and C(20, 44). Find the slope of the line joining the midpoints of $\stackrel{‾}{\mathrm{AB}}$ and $\stackrel{‾}{\mathrm{BC}}$.

 a. - $\frac{23}{6}$ b. - 6 c. - $\frac{11}{2}$ d. - $\frac{11}{3}$

#### Solution:

Midpoint of line joining (x1 , y1) and (x2 , y2) = (x1+x22, y1+y22)

Midpoint of OA = (282 , 0) = (14, 0)
[Step 1.]

Midpoint of AB = (28+402 , 202) = (34, 10)
[Step 1.]

Midpoint of BC = (20+402 , 44+202) = (30, 32)
[Step 1.]

Midpoint of OC = (202 , 442) = (10, 22)

= - 112
Slope of the line segment joining the midpoint of AB & BC = 32-1030-34 = - 224
[Slope of the line segment joining the points (x1, y1) and (x2, y2) = y2-y1x2-x1.]

13.
ABCD is a square. The coordinates of A & B are A(5, 0) and B(13, 0). ACPQ is another square with point D inside it. Find the coordinates of P and Q.

 a. (13, 8) and (5, 0) b. (5, 16) and (- 3, 8) c. (18, 0) and (8, 0) d. (21, 0) and (13, - 16)

#### Solution:

CD = QD.
Square with AC as side will have D as its point of intersection of the diagonals.

The coordinates of C and D are (13, 8) and (5, 8).
[Since ABCD is a square, AB = AD
OB - OA = 13 -5 = 8.]

= 5
x-coordinate of P = x-coordinate of A.

= 8 + 8
= 16
y-coordinate of P = AD + y-coordinate of D

= 5 - 8
= - 3
x-coordinate of Q = x-coordinate of D - QD.

y-coordinate of Q = y-coordinate of D = 8.

The coordinates of P and Q are (5, 16) and (- 3, 8).

14.
Coordinates of the vertices of a quadrilateral are A(3, 4), B(4, 6), C(3, 8) and D(2, 6). Identify the figure without drawing it.
 a. Rectangle b. Square c. Rhombus d. Parallelogram

#### Solution:

AB = 5, BC = 5, CD = 5, AD = 5
[Length of a line segment = (y2-y1)2 +(x2-x1)2 .]

Slope of AB = 6 - 44 - 3 = 2.

Slope of AD = 6 - 42 - 3 = - 2.

Slope of AM × slope of AD = 2 × - 2 = - 4.

AB and AD are not perpendicular.

ABCD is a rhombus.
[ABCD is not a square as the adjacent sides are not perpendicular.]

15.
The vertices of a quadrilateral are A(2, 5), B(6, 1), C(8, 3) and D(4, 7). Identify the quadrilateral with out making the diagram.
 a. None of these b. Rectangle c. Rhombus d. Square

#### Solution:

AB = (6-2)2+(5-1)2 = 32
[Length of line segment passing through (x2, y2) and (x1, y1) = (x2-x1)2+(y2-y1)2.]

BC = (8-6)2+(3-1)2 = 8

CD = (4-8)2+(7-3)2 = 32

DA = (2-4)2+(5-7)2 = 8

AB = CD ; BC = AD.
[ABCD is a parallelogram.]

Slope of AB = 1-56-2 = - 1
[Slope of line passing through (x2, y2) and (x1, y1) = y2-y1x2-x1.]

Slope of BC = 3-18-6 = 1

Slope of AB × slope of BC = - 1 × 1 = - 1

AB BC
[Product of slope = - 1.]

mB = 90°

Since ABCD is a parallelogram and mB = 90°, ABCD is a rectangle.

16.
ABCD is an isosceles trapezoid. What will be the coordinates of D if the trapezoid is moved towards right so that A is placed at the origin and $\stackrel{‾}{\mathrm{AB}}$ is on the x - axis? [Given A(- 8, 0), B(8, 0) and C(5, 6).]

 a. (- 3, 6) b. (3, 6) c. (- 3, - 6) d. (3, - 6)

#### Solution:

BP = 8 - 5 = 3
AQ = 8 - 5 = 3

x-coordinate of D = - 8 + 3 = - 5.

Coordinates of D are (- 5, 6).

When trapezoid moves towards right, A coincides with O.

x-coordinates are increased by 8.

New coordinates of D are ( (- 5 + 8), 6) = (3, 6)

17.
OABC is a parallelogram. Coordinates of A and C are (12, 0) and (5, 5). Find the product of the slope of the side $\stackrel{‾}{\mathrm{AB}}$ and diagonal $\stackrel{‾}{\mathrm{OB}}$.

 a. $\frac{5}{6}$ b. $\frac{5}{17}$ c. $\frac{5}{17}$ d. $\frac{17}{5}$

#### Solution:

Coordinates of A are (12, 0)

Coordinates of C are (5, 5).

OA = 12
BC = 12

Coordinates of B are ((5+12), 5) = (17, 5)
[OA = BC, OA parallel BC.]

Slope of AB = 5-017-12 = 55
[Slope of line passing through (x1 , y1) and (x2 , y2) is y2-y1x2-x1.]

Slope of OB = 5-017-0 = 517.

Product of the slopes = 5217(5) = 2585 = 5 / 17.

18.
In the quadrilateral OABC, slope of the diagonal $\stackrel{‾}{\mathrm{OB}}$ is $\frac{2}{3}$, slope of $\stackrel{‾}{\mathrm{AB}}$ is $\frac{3}{4}$. If the coordinates of A are (6, 0), then find the length of $\stackrel{‾}{\mathrm{AB}}$ rounded to the nearest whole number.

 a. 55 b. 60 c. 50 d. 65

#### Solution:

Slope of OB = 23.

Let the coordinates of B are (x, y).
23 = yx
y = 23x.
[Slope of a line passing through (x, y) and origin is yx.]

Slope of AB = 34

Slope of AB = y-0x-6 = yx-6
[Slope of line passing through (x2, y2) and (x1, y1) is y2-y1x2-x1.]

34 = yx-6
[From step 3 and step 4.]

23 xx-6 = 2x3(x-6)
[Substitute the value of y.]

34 = 2x3(x-6)
[From step 5 and step 6.]

3 × 3(x - 6) = 2x × 4

9x - 8x = 54

x = 54

y = 23×54 = 36
[Substitute x in step 2.]

The coordinates of B are (54, 36)

AB = (6 - 54)2+(0-36)2 = (-48)2+(36)2 = 3600 = 60
[Rounded to the nearest whole number.]

19.
ABCD is a rectangle. The coordinates of the vertices are given as A(4, 7), B(19, 7). The length of $\stackrel{‾}{\mathrm{AB}}$ is 15 units and $\stackrel{‾}{\mathrm{AC}}$ is 17 units. Find the slope of $\stackrel{‾}{\mathrm{BD}}$.

 a. - $\frac{15}{8}$ b. $\frac{15}{8}$ c. $\frac{8}{15}$ d. - $\frac{8}{15}$

#### Solution:

AB = 15
AC = 17
BC = 8
[BC2 = AC2 - AB2.]

Coordinates of C are (19, 7 + 8) = (19, 15)
[x-coordinates of B and C are same.]

Coordinates of D are (4, 7 + 8) = (4, 15)
[x-coordinates of A and D are same.]

Coordinates of B are B(19, 7)
[Given.]

Slope of BD = 15 - 74 - 19 = - 8 / 15

20.
Find whether the diagonals of the isosceles trapezoid intersect at right angle. [Given $a$ = 11, $b$ = 4 and $c$ = 6.]

 a. No b. Yes

#### Solution:

Coordinates of A are (11, 0)
[From the question.]

Coordinates of C are (4, 6)

x-coordinate of C = 4.

y-coordinate of C = 6.

y-coordinate of B = y-coordiante of C = 6.
[OA parallel to CB.]

x-coordinate of B = 11 - 4 = 7
[x-coordinate of C - x-coordinate of O = x-coordinate of A - x-coordinate of B.]

Slope of OB = 67.

Slope of AC = 64 - 11 = - 67

Product of the slopes = -627×7 = - 36 / 49

So, the diagonals of the isosceles trapezoid do not intersect at right angle.