Question 1 1 / -0
In the following figure, the area of the shaded region is _______.
Solution
The base of the shaded triangular region along the y-axis = 4 units.
Question 2 1 / -0
The vertices of △MNO are O (4, -8), N (-2, 0) and M (4, 6), as shown in the figure. Find the area of △MNO.
Solution
Vertices of the triangle are: M (x
1 , y
1 ) = (4, 6)
N (x
2 , y
2 ) = (-2, 0)
O (x
3 , y
3 ) = (4, -8)
Area of triangle =
=
=
= 42 square units
Question 3 1 / -0
A quadrilateral is given with the coordinates of its vertices in the figure below. Find its area.
Solution
E (x
1 , y
1 ) = (-8, -4)
F (x
2 , y
2 ) = (-6, -10)
G (x
3 , y
3 ) = (6, -4)
Area of ∆EFG =
=
=
= 42 square units
E (x
1 , y
1 ) = (-8, -4)
G (x
2 , y
2 ) = (6, -4)
D (x
3 , y
3 ) = (4, 6)
Area of ∆EGD =
=
=
= 70 square units
So, Area of EFGD = Area of ∆EFG + Area of ∆EGD
= 42 + 70 = 112 square units
Question 4 1 / -0
Which of the given statements is true for the following points?
Solution
Let the points (2, 10), (4, 6) and (-4, -22) be the vertices of a triangle.
M (x
1 , y
1 ) = (2, 10)
N (x
2 , y
2 ) = (4, 6)
O (x
3 , y
3 ) = (-4, -22)
Length of MN =
=
=
=
=
units
Length of NO =
=
=
=
units
Length of MO =
=
=
=
units
Now, MN + MO ≠ NO, MN ≠ NO + MO and MO ≠ NO.
So, the points (2, 10), (4, 6) and (−4, −22) are not collinear.
Question 5 1 / -0
If points (x, 0), (0, y) and (2, 2) are collinear, then which of the following is true?
Solution
If the given three points are collinear, then the area of the triangle formed by them will be 0.
Area of the triangle formed =
{x(y - 2) + 0(2 – 0) + 2(0 – y)} = 0
xy - 2x - 2y = 0
+
= 1
Question 6 1 / -0
Coordinates of points A, B and C are (2, 6), (4, 3) and (6, 6), respectively. What kind of a triangle is ΔABC?
Solution
Length of side AB =
=
units
Length of side BC =
=
units
Length of side AC =
= 4 units
Since two of its sides are equal, it is an isosceles triangle.
Question 7 1 / -0
The three vertices of a parallelogram ABCD are A(1, -2), B(3, 6) and C(5, 10). The co-ordinates of the fourth vertex D are
Solution
Let the fourth vertex be D(h, k).
Since the mid-points of diagonals of a || gm coincide, so we have:
=
⇒ h = 3
And
=
⇒ k = 2
Fourth vertex is D(3, 2).
Question 8 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 9 1 / -0
What is the area of the triangle formed by joining the points (a, a + 2), (a + 2, a + 2) and (a - 2, a)?
Solution
Area =
=
{a(a + 2 - a) + (a + 2)(a - a - 2) + (a - 2)(a + 2 - a - 2)}
=
{a(2) + (a + 2)(-2)} = 2 sq. units
Question 10 1 / -0
The ratio in which the line segment joining points (3, -4) and (-5, 6) is divided by the y-axis is
Solution
Let the y-axis divide the line segment in ratio of k : 1 at point P.
The coordinates of the endpoints are (3, -4) and (-5, 6).
Coordinates of P:
Since the point lies on y-axis,
= 0
3 - 5k = 0
k = 3/5
Required ratio = k : 1 = 3/5 : 1 = 3 : 5
Question 11 1 / -0
Point A (2, p) lies in the first quadrant at a distance of 3 units from point (2, 2). Find the possible value of p.
Solution
Applying distance formula,
3 = p - 2
p = 3 + 2 = 5
Question 12 1 / -0
What are the coordinates of the centroid of a triangle formed by joining points (2, 3), (3, 4) and (1, 2)?
Solution
The centroid of a triangle is the point of intersection of the three medians of the triangle (each median connecting a vertex with the midpoint of the opposite side).
So, the coordinates of the centroid of the triangle can be found by calculating the average of the abscissae and the ordinates of all three vertices of the triangle.
Abscissa of centroid =
=
= 2
Ordinate of centroid =
=
= 3
So, coordinates of the centroid are (2, 3).
Question 13 1 / -0
A line segment joining the points (m, n) and (n, m) is divided in the ratio n : m internally by a point P. What will be the ordinate of this point P?
Solution
Coordinates of the point P, which divides the line segment joining points
A(x
1 , y
1 ) and B(x
2 , y
2 ) in the ratio m
1 : m
2 (internally), are x =
,
y =
, where m
1 + m
2 0. (Section Formula)
Therefore, y =
y =
So, the ordinate is
.
Question 14 1 / -0
A triangle has a base of 12 units on the line 3x + 7y = 12. Find the area of the triangle if the third vertex is at (3, –5).
Solution
Perpendicular distance of point (3, –5) from the equation of line 3x + 7y = 12 is
.
Perpendicular distance will be
units, which is the altitude of the triangle.
Therefore, base = 12 units and altitude =
units.
Area of the triangle =
=
sq. units
Question 15 1 / -0
What are the coordinates of the third vertex of a triangle if the coordinates of its other two vertices are (-3, 2) and (2, -2), and its centroid lies at
?
Solution
Let coordinates of the third vertex be (a, b).
x
1 = a, x
2 = 2, x
3 = -3
y
1 = b, y
2 = -2, y
3 = 2
Centroid is
Coordinates of the centroid =
⇒
=
So,
And
a = -2 + 1 and b = 3(-1)
a = -1 and b = -3
The coordinates of the third vertex are (-1, -3).
Option (4) is correct.
Question 16 1 / -0
A point moves 3 units to the east from the origin, then 5 units to the south, then 10 units to the west and finally 15 units to the north. In which quadrant will it lie?
Solution
B is the final position of the point. It lies in the 2
nd quadrant.
Question 17 1 / -0
The coordinates of the vertices of a rhombus PQRS are given in the figure below. Find the area of the rhombus.
Solution
Length of diagonal PR =
=
=
units
Length of diagonal QS =
=
=
=
So, area of rhombus =
= 96 square units
Question 18 1 / -0
Danial and Rena are both at origin facing in the positive y direction. If Danial moves 3 units to the right and two units upward, then what will be the distance between Danial and Rena?
Solution
Danial moves 3 units to the right, which means 3 units along positive x-axis, and then 2 units upward, which means 2 units along positive y-axis.
Coordinates of Danial's final position = (3, 2)
Coordinates of Rena's position = (0, 0)
Distance from (3, 2) to (0, 0) =
=
units
Question 19 1 / -0
A boy was in a park. He started walking from point (1, 5), reached point (5, 0), then moved to point (1, -5) and finally stopped at point (-3, 0). Find the area of the quadrilateral formed by joining these four points.
Solution
Let the points be: P (1, 5), Q (5, 0), R (1, -5), S (-3, 0)
PQRS is a rhombus as its diagonals are perpendicular.
So, area of PQRS =
sq. units
Question 20 1 / -0
EFG is a triangle, where E is the vertex with coordinates (2, 1) and FG is the base with end coordinates (6, -2) and (8, 9). EH is an angle bisector, where H meets FG. What are the coordinates of H?
Solution
EF =
=
=
= 5
EG =
=
=
= 10
EH is an angle bisector, so
.
=
= 1 : 2
Hence, H divides FG in the ratio of 1 : 2.
So, coordinates of H are
=
=
Question 21 1 / -0
Determine whether the given statements are true (T) or false (F) and choose the correct option accordingly:
(i) Point (-5, -4) and point (-3, -2) lie in the same quadrant.
(ii) A point lying on the x-axis at a distance of -4 units from the y-axis has coordinates (0, -4).
(iii) Abscissa of a point is negative in second quadrant as well as in third quadrant.
(iv) There are more points than one that lie on both the axes.
(i) (ii) (iii) (iv) A T F F F B T F T F C F T F T D F T T T
Solution
(i) Point (-5, -4) lies in third quadrant, and point (-3, -2) also lies in third quadrant.
Question 22 1 / -0
Fill in the blanks and choose the correct option accordingly:
(i) The abscissa and the ordinate of a point B lie 4 units to the left of the abscissa and one unit above the ordinate, respectively, of point A (3, -2). Point B lies in _____ quadrant.
(ii) The abscissa and the ordinate of a point lie 0.04 unit to the left of the abscissa and 0.04 unit below the ordinate, respectively, of point A (-10.96, 0.24). Point B lies in ____ quadrant.
(iii) The abscissa and the ordinate of a point lie 20 units to the right of the abscissa and 15 units below the ordinate, respectively, of point A (-15, 0). Coordinates of point B are ______.
(iv) Rohit moves 60 units towards east, then 30 units towards west and finally 10 units towards north. Considering his initial position to be origin, the coordinates of his final position must be ______.
(i) (ii) (iii) (iv) A III II (5, -15) (30, 10) B IV II (5, 15) (-20, 0) C III IV (-5, -10) (0, 10) D II III (-5, -15) (20, -10)
Solution
(I) Coordinates of point B are (-1, -1). So, it lies in III quadrant.
Question 23 1 / -0
In the given figure, ACBD is a parallelogram with its diagonals AB and CD along the coordinate axes, AB = 18 units and CD = 14 units. Now, if S is a point which is 5 units to the right and 7 units below C, then find
(i) the sum of the abscissae of C, B and S
(ii) the sum of the ordinates of D, A and S
Choose the correct option from below:
A (i) -9 (ii) 2 B (i) -7 (ii) 2 C (i) -8 (ii) 1 D (i) -7 (ii) 1
Solution
From the given information:
Question 24 1 / -0
In the given figure, ABC is an isosceles triangle, where altitude AO is drawn from origin O, length AO = 5a units and base BC = 6b units. Find the coordinates of all its vertices.
Choose the correct option from below:
P A (0, 5a) B (6b, 0) C (0, -3b) Q A (0, 6b) B (3b, 0) C (-3b, 0) R A (0, 5a) B (-3b, 0) C (3b, 0) S A (5a, 0) B (0, 3b) C (0, 6b)
Solution
We know that AO = 5a units and BC = 6b units.
Question 25 1 / -0
Match the following:
Column I Column II (P) The area of rectangle OACB, with vertices O (0, 0), A (4, 0), B (0, 8) and C (4, 8), is (i) 30 sq. units (Q) The area of triangle ABC, with vertices A (0, 1), B (0, 13) and C (5, 5), is (ii) 32 sq. units (R) The area of triangle CAB, with vertices C (-5, 0), A (5, 0) and B (0, 8), is (iii) 40 sq. units
Choose the correct match from below:
A P - (iii) Q - (ii) R - (i) B P - (i) Q - (iii) R - (ii) C P - (ii) Q - (iii) R - (i) D P - (ii) Q - (i) R - (iii)
Solution
(P) Area of rectangle OACB = Length × Breadth = OA × OB
= 4 × 8 = 32 sq. units
(Q) Area of triangle ABC =
× 12 × 5 = 30 sq. units
(R) Area of triangle CAB =
× 10 × 8 = 40 sq. units
×
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