Question 1 1 / -0
The points (4, 4), (3, 5) and (-1, -1) form the vertices of a/an
Solution
A = (4, 4), B = (3, 5), C = (-1, -1)2 = (3 - 4)2 + (5 - 4)2 = 22 = (-1 - 3)2 + (-1 - 5)2 = 16 + 36 = 522 = (-1 - 4)2 + (-1 - 4)2 = 25 + 25 = 502 + (AC)2 = BC2
Question 2 1 / -0
Three vertices of a parallelogram are (1, 3), (2, 0) and (5, 1). Its fourth vertex is
Solution
The diagrammatic representation of the same is as shown below:
As diagonals of a paralleogram bisect,
=
x = 4
And,
y = 4
Thus, the coordinates of the 4
th vertex are (4, 4).
Thus, answer option 2 is correct.
Question 3 1 / -0
The points A(- 4, - 1), B(- 2, - 4), C(4, 0) and D(2, 3) are the vertices of a
Solution
AB =
=
=
BC =
=
=
CD =
=
=
DA =
=
=
Since AB = CD and BC = DA. But as the four sides are not equal, it is neither a rhombus nor a square.
Also, AC =
=
=
Since AB
2 + BC
2 = AC
2 , so AB and BC are perpendicular.
Hence, ABCD is a rectangle. Since, every rectangle is parallelogram. Hence, option (4) is correct.
Question 4 1 / -0
What is the area of the triangle formed by the points A (1, 2), B (-2, 3) and C (-3, -4)?
Solution
If A (x
1 , y
1 ), B (x
2 , y
2 ) and C (x
3 , y
3 ) form a triangle, then the area of
ABC =
|(x
1 y
2 - x
2 y
1 ) + (x
2 y
3 - x
3 y
2 ) + (x
3 y
1 - x
1 y
3 )|
So, the area of
ABC =
|(x
1 y
2 - x
2 y
1 ) + (x
2 y
3 - x
3 y
2 ) + (x
3 y
1 - y
3 x
1 )|
=
|(1
3 - (-2)
2) + ((-2)
(-4) - (-3) × 3) + ((-3)
2 - (-4)
1)| sq. units
=
|(3 + 4) + (8 + 9) + (-6 + 4)| sq. units
=
|7 + 17 - 2| sq. units
= 11 sq. units
Question 5 1 / -0
If the points (-6, -5), (0, b) and (2,
) are collinear, what is the value of b?
Solution
If the points A(x
1 , y
1 ), B(x
2 , y
2 ) and C(x
3 , y
3 ) are collinear, then the area of
ABC =
|(x
1 y
2 - x
2 y
1 ) + (x
2 y
3 - x
3 y
2 ) + (x
3 y
1 - x
1 y
3 )| is equal to zero.
Let the points be A(-6, -5), B(0, b) and C(2,
).
So, the area of
ABC =
|(- 6
b - 0
(- 5) + (0
- 2
b) + (2
(- 5) - (- 6)
)| = 0
|- 6b - 2b - 10 + 10| = 0
|- 8b| = 0
b = 0
Question 6 1 / -0
If the point A (a, b) divides the line segment joining the points (2, 3) and (2, 7) internally in the ratio 1 : 3, what are the coordinates of this point?
Solution
Coordinates of the point A, which divide the line segment joining the points B (x
1 , y
1 ) and C (x
2 , y
2 ) in the ratio m
1 : m
2 (internally) are x =
and y =
, m
1 + m
2 0.
a =
=
= 2
b =
=
=
a = 2 and b = 4
So, the coordinates of point A are (2, 4).
Question 7 1 / -0
If (2, 3), (3, a) and (b, – 2) are the vertices of a triangle whose centroid is (0, 0), then find the respective values of a and b.
Solution
x-coordinate of the centroid =
0 =
Solving, we get
b = - 5
Similarly,
y-coordinate of centroid =
a = - 1
Question 8 1 / -0
In the rectangular coordinate system given above, the area of triangle RST is
Solution
Area of
RST =
Length of altitude
Length of base
Length of base = (c - 1) units
Length of altitude = b units
Area of
RST = (
b
(c - 1)) sq. units
Hence, option (2) is correct.
Question 9 1 / -0
What is the ratio in which P (4, 6) divides the line joining A (-2, 3) and B (6, 7)?
Solution
Let the ratio be m
1 : m
2 .
Coordinates of P =
= (4, 6)
= 4
6m
1 - 2m
2 = 4m
1 + 4m
2 2m
1 = 6m
2 Ratio = 3 : 1
Option 1 is correct.
Question 10 1 / -0
Find the radius of a circle whose diameter has endpoints (-3, -2) and (7, 8).
Solution
Endpoints of the diameter are (-3, -2) and (7, 8),
Hence,
Diameter =
units
=
units
=
units
=
units
=
units
Radius = Diameter/2
=
units
Question 11 1 / -0
Point P(x, y) is equidistant from the points (1, 3) and (2, –1). Which of the following equations holds true?
Solution
If a point (x, y) is equidistant from points (x
1 , y
1 ) and (x
2 , y
2 ), then the distance of the point (x, y) from (x
1 , y
1 ) is equal to the distance of the point (x, y) from (x
2 , y
2 ).
So, distance of the point (x, y) from (1, 3) = Distance of the point (x, y) from (2, –1)
=
[
Distance between two points (x
1 , y
1 ) and (x
2 , y
2 ) is
]
Squaring both the sides, we get
(x – 1)
2 + (y – 3)
2 = (x – 2)
2 + (y + 1)
2 x
2 + 1 – 2x + y
2 + 9 – 6y = x
2 + 4 – 4x + y
2 + 1 + 2y [
(a
b)
2 = a
2 + b
2 2ab]
–2x – 6y + 10 = –4x + 2y + 5
–2x + 4x – 6y – 2y + 10 – 5 = 0
Or, 2x – 8y + 5 = 0
Thus, 2x – 8y = –5
Question 12 1 / -0
Point (0, y) is equidistant from the points (7, 6) and (-3, 4). What is the value of y?
Solution
If a point (x, y) is equidistant from the points (x
1 , y
1 ) and (x
2 , y
2 ), then the distance of point (x, y) from (x
1 , y
1 ) is equal to the distance of point (x, y) from (x
2 , y
2 ).
So,
the
distance of the point (0, y) from (7, 6) = Distance of the point (0, y) from (-3, 4)
Therefore,
=
Squaring both sides, we get
7
2 + (y - 6)
2 = 3
2 + (y - 4)
2 49 + y
2 + 36 - 12y = 9 + y
2 + 16 - 8y [
(a - b)
2 = a
2 + b
2 - 2ab]
-12y + 8y = 25 - 85
-4y = - 60
y =
Or y = 15
Question 13 1 / -0
The centre of a circle passing through the points (7, -5), (3, -7) and (3, 3) is
Solution
Let the points be A(3, -7), B(7, -5) and C(3, 3).
From the slope of the line,
Slope of BC is m
1 = (3 + 5)/(-4) = -2
Slope of AB is m
2 = (-5 + 7)/(7 - 3) =
As m
1 m
2 = -1, we have
AB is perpendicular to BC.
Thus, triangle ABC is a right-angled triangle, right-angled at B.
Thus, the centre of the circle passing through the three points is the midpoint of AC.
Applying the midpoint formula, the coordinates of the required point =
=
= (3, -2).
Question 14 1 / -0
If the points (2a, a), (a, 2a) and (a, a) enclose a triangle of area 2 sq. units, then the value of a is
Solution
The diagrammatic representation of the same is as shown in the diagrams below.
In either case,
Area of
a
2 units
2 = 2 units
2 or a
2 = 4 or a = ± 2
As -2 is the only option available, a = -2
Hence, answer option 1 is correct.
Question 15 1 / -0
In the figure given above, what is the area of the quadrilateral OACB?
Solution
Area of quadrilateral OACB = Area of right-angled ΔAOD - Area of ΔBCD
Area of right-angled ΔAOD = (
12
6) sq. units = 36 sq. units
Area of ΔBCD = (
8
2) sq. units = 8 sq. units
Hence, area of quadrilateral OACB = (36 - 8) sq. units = 28 sq. units
Hence, option 2 is correct.
Question 16 1 / -0
Ajay and Vijay are two local drivers who drive through different routes. Ajay drives within the city. He starts from point A(2, 2) and then moves to point B(6, 4), then to point C(10, 12), then to point D(14, 18) and then reaches his final destination at E(18, 22), where the coordinates are in km. Vijay's route is outside the city from point A to point E. What is the difference between the distances covered by them, in metres? (Consider the value of each up to two decimal places)
Solution
Distance covered by Ajay = Distance from point A to point B + Distance from point B to point C + Distance from point C to point D + Distance from point D to point E
Distance from point A to point B =
= 4.47 km
Distance from point B to point C =
= 8.94 km
Distance from point C to point D =
= 7.21 km
Distance from point D to point E =
= 5.66 km
Total distance = (4.47 + 8.94 + 7.21 + 5.66) km = 26.28 km
Distance covered by Vijay = Distance from point A to point E =
= 25.61 km
Difference between the distances covered = (26.28 – 25.61) = 0.67 km = 670 m
Question 17 1 / -0
Directions For Questions
Directions: Study the following information and answer the question that follows.
Different flowers are planted in the rows and column in a field yard for a flower show. Out of them, A, B, C and D are four flowers which are shown in the figure. All the coordinates are in metres. (Consider the value of each up to two decimal places.)
...view full instructions
What is the difference of distance between AC and AB?
Solution
According to the question:
Distance from A to C =
Distance from A to B =
So, required difference = 16.97 – 4.47 = 12.5 m
Question 18 1 / -0
Directions For Questions
Directions: Study the following information and answer the question that follows.
Different flowers are planted in the rows and column in a field yard for a flower show. Out of them, A, B, C and D are four flowers which are shown in the figure. All the coordinates are in metres. (Consider the value of each up to two decimal places.)
...view full instructions
If there is a new flower blooming at 12 units along X axis and 8 units along Y axis with respect to point A in the same field yard, then what will be the difference between the ordinates of point D and the new point formed?
Solution
According to the conditions given, the new point is formed with coordinates (16, 14).
Question 19 1 / -0
A street starts from house A at point (5, 22) and ends at house B at point (-3, 8). Another street which is perpendicular to the given street joins another house C to midpoint of the street joining houses A and B. Which of the following could be the coordinates of house C?
Solution
Midpoint of (5, 22) and (-3, 8) =
=
=
= (1, 15)
Slope of the line joining (5, 22) and (-3, 8) =
=
=
=
Slope of the line perpendicular to the given line = -1/(slope of the given line) =
Testing the slope of line joining point (1, 15) with the options:
Option 1: Slope of the line joining (3, 4) and (1, 15) =
Option 2: Slope of the line joining (5, 9) and (1, 15) =
Option 3: Slope of the line joining (-6, 19) and (1, 15) =
Option 4: Slope of the line joining (-3, 8) and (1, 15) =
Hence, (-6, 19) could be the coordinates of house C.
Question 20 1 / -0
A triangular platform is made for a flag post whose vertices are at coordinates (7, 39), (15, 9) and (4, 5). The pole is placed where the centroid of the triangle exists.
Solution
Vertices of the triangle:
A = (7, 39)
B = (15, 9)
C = (4, 5)
Area of the triangle =
=
=
=
=
=
= 181 sq units
Centroid of the triangular platform =
=
=
=
Question 21 1 / -0
A triangle has 12 units base on the line 3x + 7y = 12. If the third vertex is at point (3, - 5), find the area of the triangle.
Solution
The perpendicular distance of the vertex (3, -5) from the line is the altitude of the triangle.
∴ Altitude =
=
=
Required area of triangle =
x 12 x
=
sq. units
Question 22 1 / -0
The four vertices of a rectangle ABCD are A(6, 16), B(16, 33), C(29, 26) and D(x, y).
i. Find the area of the rectangle in square units.
ii. Find the point where diagonals of the rectangle intersect.
Solution
A diagonal divides the rectangle into two triangles of equal area.
Area of triangle ABC =
=
=
=
=
= 145.5 sq. units
Area of rectangle ABCD = 2 × area of triangle ABC = 2 × 145.5 = 291 sq. units
Diagonals of a rectangle divide each other into 2 equal parts.
So, diagonals of the rectangle intersect at the midpoint of point AC.
Midpoint of AC =
=
= (17.5, 21)
Question 23 1 / -0
The ratio of areas of triangles ABC and DEF is 1 : 3, respectively. Find the value of y.
Solution
Area of triangle ABC =
=
=
= 5 square units
We know that the ratio of areas of the triangles is 1 : 3.
So, area of triangle DEF = 15 square units
Area of triangle DEF =
15 =
15 =
30 = 4 – 4y + 24
4y = 4 + 24 – 30
4y = -2
y = -2 ÷ 4
y = -0.5
Question 24 1 / -0
The area of a rectangle ABCD is 60 square units. The coordinates of point A are (3, 4). If the perimeter of the rectangle is 34 square units, find the answers to the questions given below:
(A) Coordinates of B
(B) Coordinates of C
(C) Ratio of AB to BC
Solution
Let AB be m units and CD be n units.
According to the question,
m × n = 60
m =
--- (i)
Also, 2(m + n) = 34
2m + 2n = 34
m + n = 17
= 17
60 + n
2 = 17n
n
2 + 60 - 17n = 0
n
2 - 12n - 5n + 60 = 0
n (n - 12) - 5(n - 12) = 0
(n - 5)(n - 12) = 0
n = 5 or 12
Since AB is the largest side, n = 5 and m = 12.
(A) Therefore, coordinates of point B = (15, 4)
(B) Coordinates of point C = (15, -1)
(C) Ratio of AB to BC = 12 : 5
Question 25 1 / -0
In the given figure, the area of triangle ABC is 48 square units. The coordinates of point A are (1, 2) and those of point B are (5, -4). If D is the midpoint of CB, find the coordinates of point D.
Solution
Area of triangle ABC = 48 square units
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