MCQ 11 Mark
Two points having same abscissae but different ordinates lie on
- Ax-axis
- By-axis
- ✓a line parallel to y-axis
- Da line parallel to x-axis
Answer
View full question & answer→Correct option: C.
a line parallel to y-axis
c
Questions · Page 1 of 2
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MCQ 21 Mark
The reflection of the point $P(-4,5)$ in $y$-axis has the coordinates
- A$(-4,-5)$
- B$(4,5)$
- C$(4,-5)$
- D$(5,-4)$
Answer
View full question & answer→B.(4,5)
The reflection of point $P(-4,5)$ is its mirror image in line mirror along $y$-axis. So, $P$ and $Q$ are equidistant from $y$-axis on its opposite sides. So, $x$-coordinate of $Q$ is 4 . Also, $P$ and $Q$ have same $y$-coordinates as $P Q$ is perpendicular to $y$-axis. Hence, the coordinates of $Q$ are $(4,5)$.
The reflection of point $P(-4,5)$ is its mirror image in line mirror along $y$-axis. So, $P$ and $Q$ are equidistant from $y$-axis on its opposite sides. So, $x$-coordinate of $Q$ is 4 . Also, $P$ and $Q$ have same $y$-coordinates as $P Q$ is perpendicular to $y$-axis. Hence, the coordinates of $Q$ are $(4,5)$.
MCQ 31 Mark
The quadrilateral formed by joining the points A (0, 0), B (5, 0), C (3, 2) and D (0, 2) in order is a
- Asquare
- Bparallelogram
- ✓trapezium
- Drectangle
Answer
View full question & answer→Correct option: C.
trapezium
c
MCQ 41 Mark
The points whose abscissa and ordinate have different signs will lie in
- AI and II quadrants
- BII and III quadrants
- CI and III quadrants
- ✓II and IV quadrants
Answer
View full question & answer→Correct option: D.
II and IV quadrants
d
MCQ 51 Mark
The points (other than origin) for which abscissa is equal to the ordinate will lie in
- AI quadrant only
- BI and II quadrants
- ✓I and III quadrants
- DII and IV quadrants
Answer
View full question & answer→Correct option: C.
I and III quadrants
c
MCQ 61 Mark
The points $O(0,0), A(4,0)$ and $B(0,4)$
- Aare collinear
- Bform a scalene triangle
- Cform an equilateral triangle
- Dform an isosceles right triangle
Answer
View full question & answer→B. form a scalene triangle
Clearly, $O A=O B=4$ units and $\angle A O B=90^{\circ}$. Hence, $O, A, B$ form an isosceles right triangle.
Clearly, $O A=O B=4$ units and $\angle A O B=90^{\circ}$. Hence, $O, A, B$ form an isosceles right triangle.
MCQ 71 Mark
The point $P(5,-1)$ on reflection in $x$-axis is mapped as $Q$ and the point $Q$ on reflection in $y$-axis is mapped as $R$, the coordinates of $R$ are
- A$(-5,-1)$
- B$(-5,1)$
- C$(5,1)$
- D$(1,5)$
Answer
View full question & answer→B. The reflection of $P(5,-1)$ in $x$-axis is point $Q(5,1)$ and its reflection in $y$-axis is the point $R(-5,1)$.
MCQ 81 Mark
The point of intersect of the coordinate axes is
- Aordinate
- Babscissa
- Cquadrant
- ✓origin
Answer
View full question & answer→Correct option: D.
origin
d
MCQ 91 Mark
The perpendicular distance of the pomt $(3,-4)$ from $x$-axis, is
- A3 dmits
- B-4 units
- C4 units
- D7 units
Answer
View full question & answer→C. 4 units
The perpendicular distance of a point from $x$-axis is the absolute value of its $y$-coordinate. Hence, the distance of the point $(3,-4)$ from $x$-axis is 4 units.
The perpendicular distance of a point from $x$-axis is the absolute value of its $y$-coordinate. Hence, the distance of the point $(3,-4)$ from $x$-axis is 4 units.
MCQ 101 Mark
The perpendicular distance of the point P (4, 3) from y-axis is
- ✓4
- B3
- C5
- Dnone of these
Answer
View full question & answer→Correct option: A.
4
a
MCQ 111 Mark
The perpendicular distance of the point P (4, 3) from x-axis is
- A4
- ✓3
- C5
- Dnone of these
Answer
View full question & answer→Correct option: B.
3
b
MCQ 121 Mark
The perpendicular distance of the point $(-7,4)$ from $y$-axis is
- A7 units
- B4 units
- C11 units
- D-7 units
Answer
View full question & answer→A. 7 units
The perpendicular distance of a point from $y$-axis is the absolute value of its $x$-coordinate. Hence, the distance of the point $P(-7,4)$ from $y$-axis is 7 units.
The perpendicular distance of a point from $y$-axis is the absolute value of its $x$-coordinate. Hence, the distance of the point $P(-7,4)$ from $y$-axis is 7 units.
MCQ 131 Mark
The ordinate of any point on x-axis is
- ✓$0$
- B1
- C-1
- Dany number
Answer
View full question & answer→Correct option: A.
$0$
a
MCQ 141 Mark
The measure of the angle between the coordinate axes is
- A$0^{\circ}$
- ✓$90^{\circ}$
- C$180^{\circ}$
- D$360^{\circ}$
Answer
View full question & answer→Correct option: B.
$90^{\circ}$
b
MCQ 151 Mark
The line represented by the equation $8 x+3 y=24$ cuts the coordinate axes at $A$ and $B$. The area of the triangle $A O B$ is
- A24 sq. units
- B12 sq. units
- C48 sq. units
- D16 sq. units
Answer
View full question & answer→B. 12 sq. units
The line $8 x+3 y=24$ cuts $x$-axis at $y=0$. Putting $y=0$ in $8 x+3 y=24$, we get $x=3$. So, the line cuts $x$-axis at $A(3,0)$. Similarly, it cuts $y$-axis at $B(0,8)$. Therefore, $O A=3$ and $O B=8$. Area of $\triangle O A B=\frac{1}{2}(O A \times O B)=\frac{1}{2} \times 3 \times 8=12$ sq. units
The line $8 x+3 y=24$ cuts $x$-axis at $y=0$. Putting $y=0$ in $8 x+3 y=24$, we get $x=3$. So, the line cuts $x$-axis at $A(3,0)$. Similarly, it cuts $y$-axis at $B(0,8)$. Therefore, $O A=3$ and $O B=8$. Area of $\triangle O A B=\frac{1}{2}(O A \times O B)=\frac{1}{2} \times 3 \times 8=12$ sq. units
MCQ 161 Mark
The image of the point (-5, 7) in y-axis has the coordinates
- ✓(5,7)
- B(-5,-7)
- C(5,-7)
- D(7,-5)
Answer
View full question & answer→Correct option: A.
(5,7)
a
MCQ 171 Mark
The image of the point (3, 4) in x-axis has the coordinates
- A(-3, 4)
- ✓(3,-4)
- C(-3,-4)
- D(4,3)
Answer
View full question & answer→Correct option: B.
(3,-4)
b
MCQ 181 Mark
The image of a point $P$ under reflection in the $x$-axis has the coordinates $(7,-3)$. The coordinates of $P$ are
- A$(7,3)$
- B$(-7,3)$
- C$(-7,-3)$
- D$(-3,-7)$
Answer
View full question & answer→A. $(7,3)$
The reflection of $Q(7,-3)$ in the $x$-axis is the point $P$ itself. Hence, the coorcmates of $A$ are $(7,3)$.
The reflection of $Q(7,-3)$ in the $x$-axis is the point $P$ itself. Hence, the coorcmates of $A$ are $(7,3)$.
MCQ 191 Mark
The distance of the point P (-6, 8) from the origin is
- A6 units
- B8 units
- ✓10 units
- D14 units
Answer
View full question & answer→Correct option: C.
10 units
c
MCQ 201 Mark
The distance of the point $P(-6,8)$ from the origin is
- A6 units
- B8 units
- C14 units
- D10 units
Answer
View full question & answer→D. 10 units
Draw perpendicular PL from $P$ on $x$-axis. Clearly, $\triangle O L P$ is a right triangle with $O L=6$ units and $L P=8$ units. Applying Pythagoras theorem, we obtain$
O P^2=O L^2+L P^2 \Rightarrow O P=\sqrt{O L^2+L P^2}=\sqrt{6^2+8^2}=10 \text { units }
$
Draw perpendicular PL from $P$ on $x$-axis. Clearly, $\triangle O L P$ is a right triangle with $O L=6$ units and $L P=8$ units. Applying Pythagoras theorem, we obtain$
O P^2=O L^2+L P^2 \Rightarrow O P=\sqrt{O L^2+L P^2}=\sqrt{6^2+8^2}=10 \text { units }
$
MCQ 211 Mark
The distance of the point P (4, 3) from the origin is
- A4
- B3
- ✓5
- D7
Answer
View full question & answer→Correct option: C.
5
c
MCQ 221 Mark
The distance between the reflections of the point P (-3, 4) in the coordinate axes is
- A8 units
- B6 units
- C14 units
- ✓10 units
Answer
View full question & answer→Correct option: D.
10 units
d
MCQ 231 Mark
The distance between the points A (-5, 12) and (7, 12) is
- A5 units
- B7 units
- ✓12 units
- D17 units
Answer
View full question & answer→Correct option: C.
12 units
c
MCQ 241 Mark
The distance between the images of points P (-7, 4) and Q (7, 4) in x-axis is
- A7 units
- B8 units
- C11 units
- ✓14 units
Answer
View full question & answer→Correct option: D.
14 units
d
MCQ 251 Mark
The area of the triangle the coordinates of whose vertices are $O(0,0), A(6,0)$ and $B(0,8)$ is
- A48 sq. units
- B24 sq. units
- C14 sq. units
- D12 sq. units
Answer
View full question & answer→B. 24 sq. units
By plotting the points $O, A$ and $B$, we observe that these points are vertices of a right triangle having two legs $O A=6$ units and $O B=8$ units. $ \therefore \quad \text { Area of } \triangle O A B=\frac{1}{2}(O A \times O B)=\frac{1}{2}(6 \times 8)=24 \text { sq. units } $
By plotting the points $O, A$ and $B$, we observe that these points are vertices of a right triangle having two legs $O A=6$ units and $O B=8$ units. $ \therefore \quad \text { Area of } \triangle O A B=\frac{1}{2}(O A \times O B)=\frac{1}{2}(6 \times 8)=24 \text { sq. units } $
MCQ 261 Mark
The area of the triangle formed by the points P (0, 1), Q (0, 5) and R (3, 4) is
- A16 sq. units
- B8 sq. units
- C4 sq. units
- ✓6 sq. units
Answer
View full question & answer→Correct option: D.
6 sq. units
d
MCQ 271 Mark
The area of the triangle formed by the points A (2, 0), B (6, 0) and C (4, 6) is
- A24 sq. units
- ✓12 sq. units
- C10 sq. units
- Dnone of these
Answer
View full question & answer→Correct option: B.
12 sq. units
b
MCQ 281 Mark
The area of the triangle formed by the point P (-3, 4) and its reflections in the coordinate axes is
- ✓24 sq. units
- B48 sq. units
- C16 sq. units
- D12 sq. units
Answer
View full question & answer→Correct option: A.
24 sq. units
a
MCQ 291 Mark
The area of the figure ABCD formed by joining A (-1, 1), B (5, 1), C (5, 6) and D (-1, 6) is
- A40 sq. units
- ✓30 sq. units
- C20 sq. units
- D16 sq. units
Answer
View full question & answer→Correct option: B.
30 sq. units
b
MCQ 301 Mark
The area of the Fig. formed by joining the point $A(4,4) B(4,-4), C(-4,-4)$ and $D(-4,4)$ in order, then area of the figure formed is
- A16 sq. units
- B32 sq. units
- C64 sq. units
- D48 sq. units
Answer
View full question & answer→C. 64 sq. units
By plotting the points $A, B, C$ and $D$, we obtain a square $A B C D$ such that $A B=8$ units, $\therefore$ Area of square $A B C D=8 \times 8=64$ sq. units.
By plotting the points $A, B, C$ and $D$, we obtain a square $A B C D$ such that $A B=8$ units, $\therefore$ Area of square $A B C D=8 \times 8=64$ sq. units.
MCQ 311 Mark
The area of the Fig. formed by joining points $A(-1,5)$ and $B(-2,4)$ and their reflections in $u$-axis is
- A3 sq. units
- B6 sq. units
- C4 sq. units
- D12 sq. units
Answer
View full question & answer→A. 3 sq, units
SOLUTION Reflections of points $A(-1,5)$ and $B(-2,4)$ in $y$-axis are points $D(1,5)$ and $C(2,4)$ respectively. Clearly, $A B C D$ is a trapezium with $A D=2, B C=4$ and distance between parallel sides $A D$ and $B C$ is 1 unit. Therefore,
Area of trapezium $A B C D=\frac{1}{2}(A D+B C) \times 1$ sq. units $=\frac{1}{2}(2+4)=3$ sq. units.
SOLUTION Reflections of points $A(-1,5)$ and $B(-2,4)$ in $y$-axis are points $D(1,5)$ and $C(2,4)$ respectively. Clearly, $A B C D$ is a trapezium with $A D=2, B C=4$ and distance between parallel sides $A D$ and $B C$ is 1 unit. Therefore,
Area of trapezium $A B C D=\frac{1}{2}(A D+B C) \times 1$ sq. units $=\frac{1}{2}(2+4)=3$ sq. units.
MCQ 321 Mark
The abscissa of a point is positive in the
- AFirst and Second quadrant
- BSecond and Third quadrant
- CThird and Fourth quadrant
- ✓Fourth and First quadrant
Answer
View full question & answer→Correct option: D.
Fourth and First quadrant
d
MCQ 331 Mark
The abscissa of any point on y-axis is
- ✓$0$
- B1
- C-1
- Dany number
Answer
View full question & answer→Correct option: A.
$0$
a
MCQ 341 Mark
The abscissa and ordinate of the origin are
- ✓(0,0)
- B(1,0)
- C(0,1)
- D(1,1)
Answer
View full question & answer→Correct option: A.
(0,0)
a
MCQ 351 Mark
Signs of the abscissa and ordinate of a point in the second quadrant are respectively
- A+,+
- B-, -
- ✓-, +
- D+, -
Answer
View full question & answer→Correct option: C.
-, +
c
MCQ 361 Mark
Points (-4, 0) and (7, 0) lie
- ✓on x-axis
- By-axis
- Cin first quadrant
- DIn second quadrant
Answer
View full question & answer→Correct option: A.
on x-axis
a
MCQ 371 Mark
Points (2, 2), (3, -3, (4, -5), (-3,-4)
- Alie in II quadrant
- Blie in III quadrant
- Clie in IV quadrant
- ✓do not lie in the same quadrant
Answer
View full question & answer→Correct option: D.
do not lie in the same quadrant
d
MCQ 381 Mark
Ordinate of all points on the y-axis is
- A$0$
- B1
- C2
- ✓any number
Answer
View full question & answer→Correct option: D.
any number
d
MCQ 391 Mark
On plotting the points O (0, 0), A (3, 0), B (3, 4), C (0, 4) and joining OA, AB, BC and CO which of the following figure is formed?
- ASquare
- ✓Rectangle
- CTrapezium
- DRhombus
Answer
View full question & answer→Correct option: B.
Rectangle
b
MCQ 401 Mark
In example 15, the distance between $O$ and $R$ is
- A2 units
- B6 units
- C5 units
- D10 units
Answer
View full question & answer→D. 10 units
It is evident from Fig. that $Q R=10$ units.
It is evident from Fig. that $Q R=10$ units.
MCQ 411 Mark
If $|x|>0$ and $y<0$, then the quadrant in which the point representing $(x, y)$ can lie are
- AI, II
- BII, III
- CIII, IV
- DIV, I
Answer
View full question & answer→C. III, IV
Since $y<0$ and $x$ can be positive or negative. So, the point $(x, y)$ lies in III or IV quadrant.
Since $y<0$ and $x$ can be positive or negative. So, the point $(x, y)$ lies in III or IV quadrant.
MCQ 421 Mark
If $x>0$ and $y<0$, Ihe point $(x,-y)$ lies in
- AI quudrant
- BII quadrant
- CIII quadrant
- DIV quadrant
Answer
View full question & answer→A. I quudrant
As $y<0$, therefore $-y>0$. Thus, both the coordinates of point $(x,-y)$ are positive. Hence, it lies in the first quadrant.
As $y<0$, therefore $-y>0$. Thus, both the coordinates of point $(x,-y)$ are positive. Hence, it lies in the first quadrant.
MCQ 431 Mark
If the point $P(x, y)$ lies in the fourth quadrant, then
- A$x>y$
- B$x
- C$x>-y$
- D$y>-x$
Answer
View full question & answer→A. $x>y$
If $P(x, y)$ lies in the fourth quadrant, then $x>0$ and $y<0 \Rightarrow x>y$.
If $P(x, y)$ lies in the fourth quadrant, then $x>0$ and $y<0 \Rightarrow x>y$.
MCQ 441 Mark
If the point P (4, 2) is translated parallel to x-axis through 8 units, then the coordinates of new position of P are
- ✓(4, 10)
- B(4,6)
- C(12, 2)
- D(4,-6)
Answer
View full question & answer→Correct option: A.
(4, 10)
a
MCQ 451 Mark
If the perpendicular distance of a point P from the x-axis is 5 units and the foot of the perpendicular lies on the negative direction of x-axis, then the point Phas
- Ax-coordinate 5
- By-coordinate = 5 only
- Cy-coordinate = -5 only
- ✓y-coordinate = 5 or -5
Answer
View full question & answer→Correct option: D.
y-coordinate = 5 or -5
d
MCQ 461 Mark
If the mirror image of the point P(5, 2) in x-axis is the point Q and the image of Qin y-axis is R. Then, the coordinates of R are
- A(5,-2)
- ✓(-5,-2)
- C(-5, 2)
- D(2,5)
Answer
View full question & answer→Correct option: B.
(-5,-2)
b
MCQ 471 Mark
If the coordinates of a point $P(x, y)$ satisfy the relation $x y>0$, then $P$ may lie
- AI or II quadrant
- BII or III quadrant
- CI or III quadrant
- DI or IV quadrant
Answer
View full question & answer→C. I or III quadrant
$x y>0 \Rightarrow(x>0$ and $y>0)$ or $(x<0$ and $y<0) \Rightarrow P$ lies either in I or in III quadrant.
$x y>0 \Rightarrow(x>0$ and $y>0)$ or $(x<0$ and $y<0) \Rightarrow P$ lies either in I or in III quadrant.
MCQ 481 Mark
If points $P$ and $Q$ have coordinates $(-2,7)$ and $(-5,9)$ respectively, then the value of $($ abscissa of $P)-($ abscissa of $Q)$, is
- A3
- B-3
- C-2
- D2
Answer
View full question & answer→A. 3
$\begin{array}{l}\text { Clearly, abscissa of } P=-2 \text { and abscissa of } Q=-5 . \\ \therefore \quad(\text { abscissa of } P)-(\text { abscissa of } Q)=-2-(-5)=3\end{array}$
$\begin{array}{l}\text { Clearly, abscissa of } P=-2 \text { and abscissa of } Q=-5 . \\ \therefore \quad(\text { abscissa of } P)-(\text { abscissa of } Q)=-2-(-5)=3\end{array}$
MCQ 491 Mark
If noints $P(3,0)$ and $O(a, 0)$ are equidistant from the origin, then $a=$
- A3
- B-3
- C6
- D9
Answer
View full question & answer→B. -3
Points $P$ and $Q$ are on $x$-axis and are equidistant from the origin. So, Q Ires on $O X$ at $a$ distance of 3 units from the origin. Therefore, coordinates of $Q$ are $(-3,0)$. Hence, $a=-3$.
Points $P$ and $Q$ are on $x$-axis and are equidistant from the origin. So, Q Ires on $O X$ at $a$ distance of 3 units from the origin. Therefore, coordinates of $Q$ are $(-3,0)$. Hence, $a=-3$.
MCQ 501 Mark
A point whose abscissa is - 3 and ordinate 2 lies in
- AFirst quadrant
- ✓Second quadrant
- CThird quadrant
- DFourth quadrant
Answer
View full question & answer→Correct option: B.
Second quadrant
b