COORDINATE GEOMETRY QUESTIONS
1. Find the coordinates of the point of intersection of the medians of triangle MNO; given M = (-2, 3), N = (6, 7), O = (4, 1).
1. (5/3, 1/3)
2. (3/8, 3/11)
3. (8/3, 11/3)
4. (5/3, 17/3)
Sol: Option 3
Point of intersection of medians is called as centroid, at which each median is divided in the ratio 2 : 1. Firstly find the midpoint of NO and take it as P
Take the point C, where medians meet and it will divide the median MP in the ratio 2: 1.
2. In which quadrant does the given point (-2, 3) lies?
A - II
B - III
C - IV
D - I
Answer - A
Explanation
(-2, 3) lies in quadrant II.
3. Find the distance between the points A (-4, 7) and B (2, 5).
A - 5
B - 6
C - 6√5
D - 7
Answer - C
Explanation
AB = √(2+4)2 + (-5-7)2 = √62+ (-12)2 = √36+144 = √180
=√36*5 = 6√5 units.
4. Find the distance of the point A (6,-6) from the origin.
A - 5
B - 6
C - 6√5
D - 6√2
Answer - D
Explanation
OA = √62+ (-6)2 =√36+36 =√72 = √36*2 = 6√2 units.
5. Show that the points A(0,-2) ,B(3,1) ,C(0,4) and D(-3,1) are the vertices of a square.
A - false
B - true
Answer - B
Explanation
AB2= (3-0)2+ (1+2)2= (9+9) =18
BC2= (0-3)2+ (4-1)2= (9+9) =18
CD2= (0-3)2+ (1+2)2= (9+9) =18
DA2= (-3-0)2+ (1+2)2= (9+9) =18
∴ AB= BC=CD=DA = √18 = √9*2 = 3√2
AC2=(0-0)2+(4+2)2= (0+36) =36
BD2= (-3-3)2+ (1-1)2= (36+0) =36
∴ Diag AC = Diag BD = 6
Thus all sides are equal and the diagonals are equal.
∴ ABCD is a square.
6. Show that the points P(-4,-1), Q(-2,-4), R(4,0), and S(2,3) are the vertices of a rectangle.
A - false
B - true
Answer - B
Explanation
PQ2= (-2+4)2+ (-4+1)2= 22+ (-3)2= (4+9) =13
QR2= (4+2)2+ (0+4)2= (62+42) = (36+16) =52
RS2= (2-4)2= (3-0)2= (-2)2+32= (4+9) = 13
SP2= (2+4)2+ (3+1)2= (62+42) = (36+16) = 52
∴ PQ=RS =√13 AND QR=SP =√52
PR2= (4+4)2+ (0+1)2= (82+12) = (64+1) =65
QS2= (2+2)2+ (3+4)2= (42+72) = (16+49) =65
∴ Diag PR= Diag QS =√ 65
Thus, opposite sides are equal and diagonals are equal.
∴ ABCD is a rectangle.
7. Show that the points A (-3, 2), B (-5-5), C (2-3) and D (4, 4) are the vertices of a rhombus.
A - false
B - true
Answer - B
Explanation
AB2= (-5+3)2+ (-5-2)2= (-2)2+ (-7)2= (4+49) =53
BC2= (2+5)2+ (-3+5)2= (7)2+ (2)2= (49+4) =53
CD2= (4-2)2+ (4+3)2= (22+72) = (4+49) =53
DA2= (4+3)2+ (4-2)2= (72+22) = (49+4) =53
∴ AB=BC=CD=DA = √53
AC2= (2+3)2+ (-3-2)2= (52) + (-5)2= (25+25) = 50
BD2= (4+5)2+ (4+5)2= (92) + (92) = (81+81) =162
∴ Diag AC ≠ Diag BD
Thus all the sides are equal and diagonals are not equal.
∴ ABCD is a Rhombus.
8. Discover the region of ABC whose vertices are A (10, - 6), B (2, 5), and C (- 1, 3).
A - 49/2 sq.units.
B - 47/2 sq.units.
C - 45/2 sq.units.
D - 43/2 sq.units.
Answer - A
Explanation
Here x1=10, x2=2, X3 = -1 and y1= - 6, y2= 5, y3= 3
∴ ∆= 1/2 {X1(y2-Y3) +x2(Y3-Y1) +X3 (Y1-Y2)}
=1/2 {10(5-3) +2(3+6) - 1(- 6-5) = 1/2 (20+18+11) =49/2 sq.units.
9. Discover the estimation of h for which the focuses A (- 1, 3), B (2, h), and C (5, - 1) are collinear.
A - 1
B - 2
C - 3
D - 4
Answer - A
Explanation
Here x1=-1, x2=2, x3=5 and y1=3, y2=h and Y3=-1
Now, ∆=0 ⇒ X1(y2-Y3) +x2(Y3-Y1) +X3(Y1-Y2) = 0
⇒ -1(h+1) +2(-1-3) +5(3-h) =0
⇒ -h-1-8+15-5h=0 ⇒ 6h=6 ⇒ h=1
10. Discover the co-ordinates of the centroid of ∆ABC whose vertices are A (6, - 2) and B (4, - 3) what's more, C (- 1, - 4).
A - (-3,-3)
B - (3,3)
C - (3,-3)
D - (-3,3)
Answer - C
Explanation
The directions of the centroid are
{(6+4-1)/3, (- 2-3-4)/3} i.e. (3, - 3)
11. Discover the proportion in which the point p (2, - 5) partitions the line portion AB joining A (- 3, 5) what’s more, B (4, - 9).
A - 1:2
B - 5:2
C - 2:5
D - 2:1
Answer - B
Explanation
Let the required proportion be x:1.
At that point (4x-3/x+1, - 9 x+5/x+1) concurs with p (2, - 5)
∴ 4x-3/ ( x+1) =2 ⇒ 4x-3 = 2x+2 ⇒ 2x=5 ⇒ x=5/2
∴ required proportion is 5/2:1 i.e. 5:2
12. Discover the slop of the line whose slant is 30°?
A - 1/√3
B - 2/√3
C - 3/√3
D - 4/√3
Answer - A
Explanation
m= tan 30° = 1/√3
13.Discover the slant of the line whose slope is 1/√3
A - 30°
B - 60°
C - 80°
D - Cannot be computed with the given information
Answer - A
Explanation
tan x = 1/√3 ⇒ x=30°
14. Discover the slop of the line which goes through focuses A (- 2, 3) and B (4, - 6).
A - 3/2
B - -3/2
C - 3/4
D - 3/5
Answer - B
Explanation
Slop of AB = y2-y1/x2-x1 = - 6-3/4+2 = - 9/6 = - 3/2
15. Discover the slop of the line whose mathematical statement is 3x+4y-5 = 0.
A - 3/4
B - -3/4
C - 1/4
D - -1/4
Answer - B
Explanation
3x+4y-5 = 0 ∴ 4y=-3x+5 ∴ y=-3/4x+5/4
∴ slop = m =-3/4
16. Discover the estimation of h for which the line 2x+3y-4 = 0 and hx+6y+5 =0 are parallel.
A - 2
B - 3
C - 4
D - 5
Answer - C
Explanation
2x+3y - 4 =0 ⇒ 3y= - 2x+4 ⇒y= - 2x/3 +4/3
hx+6y+5 =0 ⇒ 6y =-hx-5 ⇒ y= - hx/6 - 5/6
The line will be parallel if - h/6 -2/3 ⇒ h= (2/3*6) = 4
∴ h=4
17. Discover the estimation of h for which the lines 5x+3y +2=0 and 3x-hy+6= 0are perpendicular to each other.
A - 2
B - 3
C - 4
D - 5
Answer - D
Explanation
5x+3y+2 =0 = -5x-2 ⇒ y= -5x/3-2/3
3x- hy+6 =0 ⇒ hy = 3x+6 ⇒ y =3x/h+6/h
The line will be perpendicular to each other if -5/3* 3/h= -1 ⇒ h=5.
Hence h= 5.
18. In the event that the separation of the point p(x, y) from A (a, 0) is a+x, then y2=?
A - 2ax
B -4ax
C - 6ax
D - 8ax
Answer : B
Explanation
(x-a) 2+(y-o) 2= a+x
⇒ (x-a) 2+y2= (a+x) 2⇒y2=(x+a) 2-(x-a) 2= 4ax ⇒ y2=4ax
19. Find the equation of the line passing through (2, -1) and parallel to the line 2x – y = 4.
1. y= 2/5x- 1
2. Y= 5x-2
3. Y= 2x-5
4. None of these
Sol: Option 3
The given line is 2x – y = 4 ⇒ y = 2x – 4 (Converting into the form of y = mx + c)
Its slope = 2. The slope of the parallel line should also be 2.
Hence for the required line
m = 2 and (x1 , y1) = (2, -1).
Equation = (y-y1) / (x/x1) = (y2-y1) / (x2-x1)
⇒ (y-y1) / (x/x1) = m
⇒ y-y1 = m (x-x1) ⇒ y - (-1) = 2(x-2)
⇒ y = 2x – 5.
20. Find the equation of straight line passing through (2, 3) and perpendicular to the line 3x + 2y + 4 = 0
1. y =5/3x- 2
2. 3Y=2x+5
3. 3Y=5x-2
4. None of these
Sol: Option 2
The given line is 3x + 2y + 4 = 0 or y = (-3x/2) - 2
Any line perpendicular to it will have slope = 2/3
Thus equation of line through (2, 3) and slope 2/3 is
(y – 3) = 2/3 (x – 2)
⇒ 3y – 9 = 2x – 4
⇒ 3y – 2x – 5 = 0.
21. If the coordinates of the centroid of a triangle are (1, 3) and two of its vertices are (–7, 6) and (8, 5), then the third vertex of the triangle is
a. (23,143)(23,143)
b. (−23,−143)(−23,−143)
c. (2, –2)
d. (–2, 2)
Sol: Let (x, y) be the coordinates of the third vertex of the triangle. Then −7+8+x3−7+8+x3 = 1 and 6+5+y3=36+5+y3=3 ⇒⇒ x + 1 = 3 and y + 11 = 9 ⇒⇒ x = 2 and y = –2 ⇒⇒ (x, y) = (2, –2)
22. The ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7) is
a. 3 : 4
b. 4 : 3
c. 2 : 3
d. 3 : 2
Sol: Let the line 3x + y – 9 = 0 divide the segment joining the points (1, 3) and (2, 7) at point P in the ratio k : 1.
The point P is (2k+1k+1,7k+3k+1)(2k+1k+1,7k+3k+1)
P lies on 3x + y – 9 = 0
3(2k+1k+1)+7k+3k+1−9=03(2k+1k+1)+7k+3k+1−9=0
⇒⇒6k + 3 + 7k + 3 – 9k – 9 = 0 4k – 3 = 0 ⇒⇒
k=34k=34
The ratio is or 3 : 4
23. Find the area of a triangle whose vertices are (1, 1), (5, 2) and (7, 4)
Solution:
Here (x1,y1)(x1,y1) = (1, 1) , (x2,y2)(x2,y2) = (5, 2) and (x3,y3)(x3,y3) = (7, 4)
Area of the triangle = 12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]
= 12(1(2−4)+5(4−1)+7(1−2))12(1(2−4)+5(4−1)+7(1−2))
= 12(−2+15−7)12(−2+15−7)
= 12(6)=312(6)=3 units
24. The distance between the points A(5, -7) and B(2, 3) is:
A.) 109
B.) 5√7
C.) √109
D.) None of these
Answer: Option C
AB2 = (2 - 5)2 + (3 + 7)2
=> (-3)2 + (10)2
=> 9 + 100 => √109
25. In which quadrant does the point(9, 0) lie?
A.) x-axis
B.) y-axis
C.) 4th
D.) None of these
Answer: Option 'A'
The point (9, 0) lies in the y-axis.
26. If the distance of the point P(x, y) from A(a, 0) is a + x, then y2 = ?
A.) 8ax
B.) 6ax
C.) 4ax
D.) 2ax
Answer: Option 'C'
√(x-a)2+(y-0)2 = a + x
= (x-a)2+y2
= (a+x)2 => y2 = (x-a)2-(x-a)2-4ax => y2 = 4ax
27. Find the area of ΔABC whose vertices are A(2, -5), B(4, 9) and (6, -1).
A.) 9 units
B.) 5 sq.units
C.) 7 sq.units
D.) 6 sq.units
Answer: Option 'D'
Here, x1 = 2, x2 = 4, x3 = 6 and y1 = -5, y2 = 9, y3 = -1
= 1/2 [2(9+1) + 4(-1+5) + 6(5-9)]
= 1/2 [2(10) + 4(4) + 6(-4)]
= 1/2 [20 + 16 - 24]
= 1/2 [12]
= 6 sq.units
28. Find the distance of the point A(4, -2) from the origin.
A.) 4√5 units
B.) 2√5 units
C.) 5√2 units
D.) 7√2 units
Answer: Option 'B'
OA = √4 - 02+(-2 - 0)2 = √16+4 = √20 = √4 × 5 = 2√5 units
29. Find the distance between the points A(-4, 7) and B(2, -5).
A.) 8√5 Units
B.) 6√5 Units
C.) 6√4 Units
D.) None of these
Answer: Option 'B'
AB = √(2+4)2 + (-5-7)2
= √62 + (-12)2
= √36+144 = √180
=√36 × 5 = 6√5 units.
30. Find the value of k for which the points A(-2, 5), B(3, k), and C(6, 1) are collinear.
A.) 5
B.) 4
C.) 7
D.) 1
Answer: Option 'D'
x>1 = -2, x>2 = 3, x>3 = 6 and y>1 = 5, y>2 = k, y>3 = -1
Now Δ = 0 <=> -2(k + 1) + 3(-1 + 5) + 6(5 - k) = 0
<=> -2 (k+1) + 3(4) + 6(5-k) = 0
<=> -2k-2 + 12+30 -6k = 0
<=> 40 - 8k = 0
<=> -8k = -40
<=> k = 5.
31. Find the area of Δ ABC whose vertices are A(9, -5), B(3, 7) and (-2, 4).
A.) 29 units
B.) 35.9 sq.units
C.) 39 sq.units
D.) 39.5 sq.units
Answer: Option 'C'
Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4
= 1/2 [9(7-4) + 3(4+5) + (-2)(-5-7)]
= 1/2 [9(3) + 3(9) - 2(-12)]
= 1/2 [27 + 27 + 24]
= 1/2 [78]
= 39 sq.units
32.If for a line m = tanϑ > 0, then
A.) ϑ is acute
B.) ϑ is obtuse
C.) ϑ = 90°
D.) ϑ = 60°
Answer: Option 'A'
m = tanϑ < 0 => is acute
33. Find the vertices of a triangle are A(2, 8), B(-4, 3) and (5, -1). The area of ΔABC is:
A.) 5 1/2 sq.units
B.) 2 1/3 sq.units
C.) 2 1/2 sq.units
D.) None of these
Answer: Option 'C'
Here, x1 = 2, x2 = -4, x3 = 5 and y1 = 8, y2 = 2, y3 = -1
= 1/2 [2(3+1) - 4(-1-8) + 5(8-3)]
= 1/2 [2(4) + 4(-9) + 5(5)]
= 1/2 [16 - 36 + 25]
= 1/2 [5]
= 2 1/2 sq.units
34. In which quadrant does the point(1, 5) lie?
A.) 1st
B.) 2nd
C.) 3rd
D.) 4th
Answer: Option 'A'
The point (1, 5) lies in 1st quadrant.
35. In which quadrant does the point(0, 9) lie?
A.) x-axis
B.) y-axis
C.) 3rd
D.) None of these
Answer: Option 'B'
The point (0, 9) lies in x-axis.
36. The points A(-4, 0), B(1, -4), and C(5, 1) are the vertices of
A.) An isosceles right angled triangle
B.) An equilateral triangle
C.) A scalene triangle
D.) None of these
Answer: Option 'A'
AB2 = (1 + 4)2 + (-4 - 0)2
= 25 + 16 = 41,
BC2 = (5 - 1)2 + (1 + 4)2 = 42 + 52
= 16 + 25 = 41
AC2 = (5 + 4)2 + (1 - 0)2
= 81 + 1 = 82
AB = BC and AB2 = BC2 = AC2
ΔABC is an isosceles right angled triangle
37. The coordinates of the endpoints of a diameter AB of a circle are A(-6, 8) and B(-10, 6). Find the coordinates of its center.
A.) -8, 7
B.) -7, 8
C.) 7, -8
D.) -8, -7
Answer: Option 'A'
The center O is the midpoint of AB.
Co-ordinates of O are [(-6-10)/2, (8+6)/2]
= -16/2, 14/2
= (-8, 7)
38. The endpoints of a line segment AB are A(-6, 4) and B(12,24). Its midpoint is:
A.) (14, 4)
B.) (-4, 14)
C.) (3, 14)
D.) (-3, 14)
Answer: Option 'D'
Midpoint is C((-6+12)/2, (4+24)/2)
(-6/2, 28/2) = (-3, 14)
39. Find the coordinates of a point P which divides the join of A(5, -4) and B(10, 8) in the ratio 3 : 2.
A.) 9, 5
B.) 7, 8
C.) 8, 7
D.) 9, -7
Answer: Option 'C'
Required point is P = (mx2 + nx1)/m+n, (my2 + ny1)/m+n
P((3(10) + 2(5))/5, (3(15) + 2(-5))/5)
=(30 + 10)/5, (45-10)/5
= P(8, 7)
40. Find the value of k for which the points A(-2, 6), B(5, k), and C(8, 3) are collinear.
A.) 5
B.) 4
C.) 7
D.) 1
Answer: Option 'A'
x1 = -2, x2 = 5, x3 = 8 and y1 = 6, y2 = k, y3 = 3
Now Δ = 0 <=> -2(k - 3) + 5(3 - 6) + 8(6 - k) = 0
<=> -2 (k-3) + 5(-3) + 8(6-k) = 0
<=> -2k + 6 - 15 + 48 - 8k = 0
<=> 39 - 10k = 0
<=> k = -(39/10)
<=> k = 3.9
41. A point C divides the join of A(2,5) and B(3,9) in the ratio 3 : 4. The coordinates of C are:
A.) (15/7, 46/7)
B.) (15/7, 47/7)
C.) (15/7, 40/7)
D.) None of these
Answer: Option 'B'
x1=2, x2=3 and y1=3, y2=7
= [(3 × 3 + 4 × 2)/7, (3 × 9 + 4 × 5)/7]
= 9+6/7, 27+20/7
= (15/7, 47/7)
42.P is a point on the x-axis at a distance of 4 units from the y-axis to its right. The coordinates of P are:
A.) (4, 0)
B.) (0, 4)
C.) (4, 4)
D.) (-4, 4)
Answer: Option 'A'
The coordinates of P are A(4, 0)
43. If the points A(1, 2), B(2, 4), and C(k, 6) are collinear, then k =?
A.) 3
B.) -3
C.) 4
D.) -4
Answer: Option 'B'
x1 = 1, x2 = 2, x3 = k and y1 = 2, y2 = 4, y3 = 6
= 1(4 - 6) + 2 (6 - 2) + k(2 - 4)
= -2 + 12 - 4 + 2k - 4k = 0
= 6 - 2k = 0
= -2k = 6
= k = -3.
44. Find the vertices of the triangle are A(2, 8), B(-4, 3) and (5, -1). The area of ΔABC is:
A.) 5 1/2 sq.units
B.) 2 1/3 sq.units
C.) 2 1/2 sq.units
D.) None of these
Answer: Option 'C'
Here, x1 = 2, x2 = -4, x3 = 5 and y1 = 8, y2 = 2, y3 = -1
= 1/2 [2(3+1) - 4(-1-8) + 5(8-3)]
= 1/2 [2(4) + 4(-9) + 5(5)]
= 1/2 [16 - 36 + 25]
= 1/2 [5]
= 2 1/2 sq.units
45. The coordinates of the endpoints of a diameter AB of a circle are A(-6, 8) and B(-10, 6). Find the coordinates of its center.
A.) -8, 7
B.) -7, 8
C.) 7, -8
D.) -8, -7
Answer: Option 'A'
The center O is the midpoint of AB.
Co-ordinates of O are [(-6-10)/2, (8+6)/2]
= -16/2, 14/2
= (-8, 7)
46. Find the area of ΔABC whose vertices are A(9, -5), B(3, 7), and (-2, 4).
A.) 29 units
B.) 35.9 sq.units
C.) 39 sq.units
D.) 39.5 sq.units
Answer: Option 'C'
Here, x1 = 9, x2 = 3, x3 = -2 and y1 = -5, y2 = 7, y3 = 4
= 1/2 [9(7-4) + 3(4+5) + (-2)(-5-7)]
= 1/2 [9(3) + 3(9) - 2(-12)]
= 1/2 [27 + 27 + 24]
= 1/2 [78]
= 39 sq.units
47.If for a line m = tanϑ < 0, then
A.) ϑ is acute
B.) ϑ is obtuse
C.) ϑ = 90°
D.) ϑ = 60°
Answer: Option 'B'
m = tanϑ < 0 => is obtuse
48. Find the distance between the points A(-4, 7) and B(2, -5).
A.) 8√5 Units
B.) 6√5 Units
C.) 6√4 Units
D.) None of these
Answer: Option 'B'
AB = √(2+4)2 + (-5-7)2
= √62 + (-12)2
= √36+144 = √180
=√36 × 5 = 6√5 units.
49. The distance between the points A(5, -7) and B(2, 3) is:
A.) 109
B.) 5√7
C.) √109
D.) None of these
Answer: Option 'C'
AB2 = (2 - 5)2 + (3 + 7)2
=> (-3)2 + (10)2
=> 9 + 100 => √109
50. In which quadrant does the point(9, 0) lie?
A.) x-axis
B.) y-axis
C.) 4th
D.) None of these
Answer: Option 'A'
The point (9, 0) lies in the y-axis.
51. The points A(0, 6), B(-5, 3) and C(3, 1) are the vertices of a triangle which is ?
A.) equilateral
B.) right angled
C.) isosceles
D.) scalene
Answer: Option 'C'
AB2= (-5 - 0)2 + (-3 - 0)2 = 16 + 9 = 25
BC2 = (3 + 5)2 + (1-3)2 = 82 + (-2)2 = 64 + 4 = 68
AC2 = (3 - 0)2 + (1 - 6)2 = 9 + 25 = 34.
AB = AC. ==> ΔABC is isosceles.
52. A point C divides the join of A(2,5) and B(3,9) in the ratio 3 : 4. The co-ordinates of C are:
A.) (15/7, 46/7)
B.) (15/7, 47/7)
C.) (15/7, 40/7)
D.) None of these
Answer: Option 'B'
x1=2, x2=3 and y1=3, y2=7
= [(3 × 3 + 4 × 2)/7, (3 × 9 + 4 × 5)/7]
= 9+6/7, 27+20/7
= (15/7, 47/7)
53. If the points A(1, 2), B(2, 4), and C(k, 6) are collinear, then k =?
A.) 3
B.) -3
C.) 4
D.) -4
Answer: Option 'B'
x1 = 1, x2 = 2, x3 = k and y1 = 2, y2 = 4, y3 = 6
= 1(4 - 6) + 2 (6 - 2) + k(2 - 4)
= -2 + 12 - 4 + 2k - 4k = 0
= 6 - 2k = 0
= -2k = 6
= k = -3.
54. P is a point on the x-axis at a distance of 4 units from the y-axis to its right. The coordinates of P are:
A.) (4, 0)
B.) (0, 4)
C.) (4, 4)
D.) (-4, 4)
Answer: Option 'A'
The coordinates of P are A(4, 0)
55. Find the distance of the point A(4, -2) from the origin.
A.) 4√5 units
B.) 2√5 units
C.) 5√2 units
D.) 7√2 units
Answer: Option 'B'
OA = √4 - 02+(-2 - 0)2 = √16+4 = √20 = √4 × 5 = 2√5 units
56. In which quadrant does the point(-4, -7) lie?
A.) 1st
B.) 2nd
C.) 3rd
D.) 4th
Answer: Option 'C'
The point (-4, -7) lies in 3rd quadrant.
57. The vertices of a quadrilateral ABCD are A(0, 0), B(3,3), C(3, 6), and D(0, 3). Then, ABCD is a
A.) square
B.) parallelogram
C.) rectangle
D.) rhombus
Answer: Option 'B'
AB2 = (3-0)2 + (3-0)2 = 18
BC2 = (3-3)2 + (6-3)2 = 9
CD2 = (0-3)2 + (3-6)2 =18
AD2 = (0-0)2 + (3-0)2 = 9
AB = CD = √18 => 3√2,
BC = AD = √9
AC2 = (3-0)2 + (6-0)2 = 9 + 36 = 45
BD2 = (0-3)2 + (3-3)2 = 9 + 0 = 9
AC ≠ BD
ABCD is a parallelogram.
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