M.C.Q (1 Marks)

MCQ 11 Mark

If A(4, 2), B(6, 5) and C(1, 4) be the verteces of $\triangle\text{ABC}$ and AD is a median, then the coordinates of D are:

A
$\Big(\frac{5}{2},3\Big)$
B
$\Big(5,\frac{7}{2}\Big)$
✓
$\Big(\frac{7}{2},\frac{9}{2}\Big)$
D
$\text{none of these}$

Answer

Correct option: C.

$\Big(\frac{7}{2},\frac{9}{2}\Big)$

Since D is the median on the side BC, D is the mid point formula, we get$\text{D}=\Big(\frac{6+1}{2},\frac{5+4}{2}\Big)$
$\text{D}=\Big(\frac{7}{2},\frac{9}{2}\Big)$

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MCQ 21 Mark

If the points A(2, 3), B(5 , k) and C(6, 7) are collinear then:

A
$\text{k}=4$
✓
$\text{k}=6$
C
$\text{k}=\frac{-3}{2}$
D
$\text{k}=\frac{11}{4}$

Answer

Correct option: B.

$\text{k}=6$

The given points are A (2, 3), B(5, k) and C(6, 7) are collinear.$\therefore(\text{x}_1=2,\text{y}_1=3),(\text{x}_2=5,\text{y}_2=\text{k)}$and $(\text{x}_3=6,\text{y}_3=7)$
The given points are collinear.
$\Rightarrow\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)=0$
$\Rightarrow2(\text{k}-7)+5(7-3)+6(3-\text{k)}=0$
$\Rightarrow5\text{k}-14+20+18-6\text{k}=0$
$\Rightarrow4\text{k}=24$
$\Rightarrow\text{k}=6$

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MCQ 31 Mark

Two vertices of $\triangle\text{ABC}$ are A(-1, 4) and B(5, 2) and its centroid is G(0, -3). Then, the coordinates of C are:

A
(4, 3)
B
(4, 15)
✓
(-4, -15)
D
(-15, -4)

Answer

Correct option: C.

(-4, -15)

Let the third vertex have coordinates D (x,y).
The centroid of a triangle is given by
$\text{C}(\text{x,y}) = \Big(\frac{\text{x}_1 + \text{x}_2 + \text{x}_3}{3},\frac{\text{y}_1 + \text{y}_2 + \text{y}_3}{3}\Big)$
$\Rightarrow \text{G}(0,-3) = \Big(\frac{-1 +5 \text{ x}}{3},\frac{4+2+\text{ y}}{3}\Big)$
$\Rightarrow \text{G}(0,-3) = \Big(\frac{4+\text{ x}}{3},\frac{4+2+\text{ y}}{3}\Big)$
$\Rightarrow \frac{4+\text{ x}}{3} = 0 $ and $\frac{6 + \text{ y}}{3} = -3$
$\Rightarrow 4+\text{x} - 0 $ and $ 6+\text{ y} = -9$
$\Rightarrow \text{x} = -4 $ and$\text{ y} = -15$
So, the coordinates of the third vertex are (-4,-15)

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MCQ 41 Mark

If A(-1, 0), B(5, -2) and C(8, 2) are the vertices of a $\triangle\text{ABC}$ then its centroid is:

A
(12, 0)
B
(6, 0)
C
(0, 6)
✓
(4, 0)

Answer

Correct option: D.

(4, 0)

Let the centrcid have coordinates C(x,y).
The centroid of a triangle is given by
$\text{C}(\text{x,y}) = \Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$
$\Rightarrow \text{C}\text{(x,y)} = \Big(\frac{-1+5+8}{3},\frac{0-2+2}{3}\Big)$
$\Rightarrow \text{C}(\text{x,y}) = (4,0)$

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MCQ 51 Mark

In the given figure P(5, -3) and Q(3, y) are the points of teisection of the line segment joining A(7, -2) and B(1, -5). Then y equals:

A
$2$
B
$4$
✓
$-4$
D
$\frac{-5}{2}$

Answer

Correct option: C.

$-4$

Since P and Q are the points of trisection.
This means AP = PQ = QB.
So, Q divides AB in the ratio 2 : 1 and let the coordinates of Q be (x, y).
We know that, the centre is the mid-point of the diameter.
Using the section formula, we get
$(3,\text{y})=\Big(\frac{2(1)+1(7)}{2+1},\frac{2(-5)+1(-2)}{2+1}\Big)$
$\Rightarrow\ \text{y}=\frac{2(-5)+1(-2)}{2+1}$
$\Rightarrow\ \text{y}=-4$

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MCQ 61 Mark

AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of each of its diagonals is:

A
$5\text{ units}$
B
$3\text{ units}$
C
$4\text{ units}$
✓
$\sqrt{34}\text{ units}$

Answer

Correct option: D.

$\sqrt{34}\text{ units}$

The diagonal $=\sqrt{(0-5)^2+(3-0)^2}$

$=\sqrt{5^2+3^2}$

$=\sqrt{25+9}$

$=\sqrt{34}\text{ units}$

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MCQ 71 Mark

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are:

✓
(-6, 7)
B
(6, -7)
C
(4, 2)
D
(5, 3)

Answer

Correct option: A.

(-6, 7)

Let the coordinates of the other end of the diameter be (x, y).We know that, the centre is the mid-point of the diameter.
Using the mid-point formula, we get:
$(-2,5)=\Big(\frac{2+\text{x}}{2},\frac{3+\text{y}}{2}\Big)$
$\Rightarrow\ \frac{2+\text{x}}{2}=-2$ and $\frac{3+\text{y}}{2}=5$
⇒ 2 + x = -4 and 3 + y = 10
⇒ x = -6 and y = 7
So, the coordinates of the oyter end are (-6, 7).

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MCQ 81 Mark

The points P(0, 6), Q(-5, 3) and R(3, 1) are the vertices of a triangle, which is:

A
Equilateral.
✓
Isosceles.
C
Scalene.
D
Right-angled.

Answer

Correct option: B.

Isosceles.

$\text{PQ}=\sqrt{(0+5^2+(6-3)^2}$$=\sqrt{25+9}=\sqrt{34}\text{ units}$
$\text{QR}=\sqrt{(-5-3)^2+(3-1)^2}$
$=\sqrt{64+4)}=\sqrt{68}\text{ units}$
$\text{PR}=\sqrt{(0-3)^2+(6-1)^2}$
$=\sqrt{9+25}=\sqrt{34}\text{ units}$
Since $\text{PQ}=\text{PR}=\triangle\text{ABC}$ is an isosceles triangle.
Also, $\text{PQ}^2=34,+\text{ PR}^2=34$ and $\text{QR}^2=68$
Clearly, $\text{PQ}^2+\text{PR}^2=\text{QR}^2,$
and so, $\triangle\text{PQR}$ is a rights-angled triangle.
Note: the answer could be (b) as well as (d), since the triangle is an isosceles right-angled triangle.

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MCQ 91 Mark

If the distance between the points A(4, p) and B(1, 0) is 5 then:

A
$\text{p} = 4\text{ only}$
B
$\text{p} = -4\text{ only}$
✓
$\text{p} = \pm4$
D
$\text{p} = 0$

Answer

Correct option: C.

$\text{p} = \pm4$

Using distance formula, we get
$\text{AB}=\sqrt{(4-1)^2+(\text{p}-0)^2}$
$\Rightarrow5=\sqrt{3^2+\text{p}^2}$
$\Rightarrow25=3^2+\text{p}^2$
$\Rightarrow25=9+\text{p}^2$
$\Rightarrow\text{p}^2=16$
$\Rightarrow\text{p}=\pm4$

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MCQ 101 Mark

Which point on $x-$axis is equidistant from the points $A(7, 6)$ and $B(-3, 4)?$

A
$(0, 4)$
B
$(-4, 0)$
✓
$(3, 0)$
D
$(0, 3)$

Answer

Correct option: C.

$(3, 0)$

Let the coordinates of the point be $P ( x , 0)$
Since the point lies on the $x -$axis. $P$ is equidistant from $A (7,6)$ and $B (-3,4)$ Using the distance from, we get
$AP^2=BP^2$
$\Rightarrow(7-x)^2+(6-0)^2=(-3-x)^2+(4-0)^2$
$\Rightarrow 49-14 x+x^2+36=9+6 x+x^2+16$
$\Rightarrow 85-14 x=6 x+25$
$\Rightarrow 20 x=60$
$\Rightarrow x=3$
Thus, the point $(3,0)$.

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MCQ 111 Mark

The point P which divides the line segment joining the points A(2, -5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant.

A
I
B
II
C
III
✓
IV

Answer

Correct option: D.

P divides AB in the ratio 2 : 3.Let the coordinates of P be (x, y).
We know that, the centra is the mid-point of the diameter.
Using the section formula, we get:
$(\text{x},\text{y})=\Big(\frac{2(5)+3(2)}{2+3},\frac{2(2)+3(-5)}{2+3}\Big)$
$\Rightarrow\ (\text{x},\text{y})=\Big(\frac{16}{5},\frac{-11}{5}\Big)$
We know that, points of the from (a, -b) lie in quadrant IV.

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MCQ 121 Mark

If the points A(x, 2), B(-3, -4) and C(7, -5) are collinear then the value of x is:

✓
-63
B
63
C
60
D
-60

Answer

Correct option: A.

-63

The given points are $\text{A} (\text{x},2) \text{B} (-3,-4) $ and $ \text{C} (7,-5)$ are cdlinear.$\therefore (\text{x}_1=\text{x},\text{ y}_1 =2) (\text{x}_2=-3,\text{ y}_2=-4), \text{ y}=-5)$ and $(\text{x}_2 =7,\text{y}_2=5)$
The given points are collinear.
$\Rightarrow \text{x}_1 (\text{y}_2 -\text{y}_2) + \text{x}_2 (\text{y}_2 -\text{y}_1) +\text{x}_2 (\text{y}_1 -\text{y}_2) =0$
$\Rightarrow \text{x}(-4+5)-3(-5-2)+7(2+4)=0$
$\Rightarrow \text{x} + 21 + 42 = 0$
$\Rightarrow \text{x} = - 63$

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MCQ 131 Mark

If R(5, 6) is the mid-point of the line segment AB joining the points A(6, 5) and B(4, 4), the y equals.

A
5
✓
7
C
12
D
6

Answer

Correct option: B.

Given that R is the mid-point of the line segment AB.The y-coordinate of $\text{R}=\frac{5+\text{y}}{2}$
$\Rightarrow6=\frac{5+\text{y}}{2}$
$\Rightarrow 12 = 5 +\text{y}$
$\Rightarrow \text{y}=7$

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MCQ 141 Mark

If $\text{P}\Big(\frac{a}{2},4\Big)$ is the mid-point of the line segment joining the points A(-6, 5) and B(-2, 3) then the value of a is:

✓
-8
B
3
C
-4
D
4

Answer

Correct option: A.

-8

Given that P $\Big(\frac{\text{a}}{2},4\Big) $is the mid -point of the line segment joiningthe points A (-6, 5) and B (-2, 3).
so, $\frac{a}{2}=\frac{-6-2}{2}$
$\Rightarrow\text{a}=-8$

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MCQ 151 Mark

In what ratio does the x-axis divide the join of A(2, -3) and B(5, 6)?

A
2 : 3
B
3 : 5
✓
1 : 2
D
2 : 1

Answer

Correct option: C.

1 : 2

Let the x-axis cut AB at the point P(x, 0) in the ratio k : 1.Then, using section formula, we get
$\frac{6\text{k}-3}{\text{k}+1}=0$
$\Rightarrow\ 6\text{k}-3=0$
$\Rightarrow\ \text{k}=\frac{1}{2}$
So, the required ratio is $\frac{1}{2}:1,$ that is 1 : 2.

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MCQ 161 Mark

The point on x-axis which is equidistant from points A(-1, 0) and B(5, 0) is:

A
(0, 2)
✓
(2, 0)
C
(3, 0)
D
(0, 3)

Answer

Correct option: B.

(2, 0)

Since the point lies on the x-axis, let the point be Pand its coordinates be (x, 0).
Given that the point is equidistant from the points A and B.
$\Rightarrow\text{PA}=\text{PB}$
$\Rightarrow\sqrt{\text{(x}+1)^2}=\sqrt{\text{(x}-5)^2}$
$\Rightarrow(\text{x}+1)^2=(\text{x}-5)^2$
$\Rightarrow\text{x}^2+2\text{x}+1=\text{x}^2-10\text{x}+25$
$\Rightarrow2\text{x}+1=-10\text{x}+25$
$\Rightarrow12\text{x}=24$
$\Rightarrow\text{x}=2$
Hence, the point is (2,0).

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MCQ 171 Mark

The points A(-4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is:

✓
Isosceles
B
Equilateral
C
Scalene
D
Right-angled

Answer

Correct option: A.

Isosceles

$\text{AC} = \sqrt{(4-4)^2 + (0-0)^2}$
$ = \sqrt{64} = 8 \text{ units}$
$\text{BC} = \sqrt{(4-0)^2 + (0-3)^2} $
$= \sqrt{16+9} = \sqrt{25} = 5 \text{ units}$
$\text{AC} = \sqrt{(-4-0)^2+)(0-3)^2} $
$= \sqrt{16+9} = \sqrt{25} = 5\text{ units}$
Since $\text{BC} = \text{AC}, \triangle\text{ABC}$ is an isosceles triangle.

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MCQ 181 Mark

The coordinates of the point P dividing the line segment joining the points A(1, 3) and B(4, 6) in the ratio 2 : 1 is:

A
(2, 4)
✓
(3, 5)
C
(4, 2)
D
(5, 3)

Answer

Correct option: B.

(3, 5)

Let the coordinates of P be (x, y).Using the section formula, we get:
$\text{P(x,y)}=\Big(\frac{2(4)+1(1)}{2+1},\frac{2(6)+1(3)}{2+1}\Big)$
$=\Big(\frac{9}{3},\frac{15}{3}\Big)$
$=(3,5)$
So, the coordinates of P are (3, 5).

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MCQ 191 Mark

The area of $\triangle\text{ABC}$ with vertice $A(a, 0), O(0, 0)$ and $B(0, b)$ in square units is:

A
$\text{ab}$
✓
$\frac{1}{2}\text{ab}$
C
$\frac{1}{2}\text{a}^2\text{b}^2$
D
$\frac{1}{2}\text{b}^2$

Answer

Correct option: B.

$\frac{1}{2}\text{ab}$

The given points are $A (a,0), 0(0, 0)$ and $B (0, b).$
$\therefore\left( x _1= a , y _1=0\right),\left( x _2=0, y _2=0\right)$ and $\left( x _2=0, y _2= b \right)$
The area of the tringle
$= \frac{1}{2}|\text{x}_1 (\text{y}_2-\text{y}_3) + \text{x}_2 (\text{y}_3-\text{y}_1) + \text{x}_3 (\text{y}_1 -\text{y}_2)|$
$=\frac{1}{2}|\text{a}(0-\text{b})+0(\text{b}-0)+0(0-0)$
$=\frac{1}{2}|-\text{ab}|$
$=\frac{1}{2}\text{ab}\text{ sq.units}$

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MCQ 201 Mark

The line 2x + y - 4 = 0 divides the line segment joining A(2, -2) and B(3, 7) in the ratio:

A
2 : 5
✓
2 : 9
C
2 : 7
D
2 : 3

Answer

Correct option: B.

2 : 9

Let the required ration be k : 1, and let P be the point of division.Using section formula, we get:
The point od division to be P$\Big(\frac{3\text{k}+2}{\text{k}+1},\frac{7\text{k}-2}{\text{k}+1}\Big).$
Since the point lies on the line 2x + y - 4 = 0, The point satisfies the equation of given line.
$\Rightarrow\ 2\Big(\frac{3\text{k}+2}{\text{k}+1}\Big)+\frac{7\text{k}-2}{\text{k}+1}-4=0$
⇒ 2(3k + 2) + 7k - 2 -4(k + 1) = 0
⇒ 6k + 4 + 7k - 2 - 4k - 4 = 0
⇒ 9k = 2
$\Rightarrow\ \text{k}=\frac{2}{9}$
So, the ratio is 2 : 9.

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MCQ 211 Mark

The distance of the point P(-6, 8) from the origin is:

A
$8$
B
$2\sqrt{7}$
C
$6$
✓
$10$

Answer

Correct option: D.

$10$

The distance of the point P(-6, 8) from the orgin (0, 0)
$=\sqrt{(-6)^2+8^2}$
$=\sqrt{36+36}$
$=\sqrt{100}$
$=10\text{ units}$

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MCQ 221 Mark

The area of $\triangle\text{ABC}$ with vertices A(3, 0), B(7, 0) and C(8, 4) is:

A
14 sq.units
B
28 sq.units
✓
8 sq.units
D
6 sq.units

Answer

Correct option: C.

8 sq.units

The given points are A (3, 0), B (7, 0) and C (8, 4).
$\therefore(\text{x}_1=3,\text{y}_1=0),(\text{x}_2=7,\text{y}_2=0)$and $(\text{x}_1=8,\text{y}_3=4)$
Area of $\triangle\text{ABC}$
$=\frac{1}{2}\mid\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\mid$
$=\frac{1}{2}\mid3(0-4)-7(4-0)+8(0-0)\mid$
$=\frac{1}{2}\mid-12+28\mid$
$=\frac{1}{2}\mid16\mid$
$=8\text{ sq}.\text{units}$

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MCQ 231 Mark

If the point C(k, 4) divides the join of the points A(2, 6) and B(5, 1) in the ratio 2 : 3 then the value of k is:

A
$16$
B
$\frac{28}{5}$
✓
$\frac{16}{5}$
D
$\frac{8}{5}$

Answer

Correct option: C.

$\frac{16}{5}$

By Section Formula,
The x-coordinate of $\text{C}=\frac{2(5)+3(2)}{2+3}$
$\Rightarrow\text{k}=\frac{16}{5}$

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MCQ 241 Mark

If the points A(1, 2), O(0, 0) and C(a, b) are collinear then:

A
a = b
B
a = mb
✓
2a = b
D
a + b = 0

Answer

Correct option: C.

2a = b

The given points are A (1,2) O(0,0) and C(a, b) are collinear.$\therefore(\text{x}_1=1,\text{y}_1=2),(\text{x}_2=0,\text{y}_2=0)$ and $(\text{x}_1=\text{a},\text{y}_1=\text{b})$
The given points are collinear.
$\Rightarrow\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)=0$
$\Rightarrow1(0-\text{b})+0+\text{a}(2-0)=0$
$\Rightarrow-\text{b}+2\text{a}=0$
$\Rightarrow2\text{a}=\text{b}$

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MCQ 251 Mark

The area of a triangle with vertices $A(5, 0), B (8, 0)$ and $C(8, 4)$ in square units is:

A
$20$
B
$12$
✓
$6$
D
$16$

Answer

Correct option: C.

$6$

The given points are $A (5,0), B (8, 0)$ and $C (8, 4).$
$\therefore\left( x _1=5, y _1=0\right),+\left( x _2=8, y _2=0\right)$ and $\left( x _2=8, y _2=4\right)$
The area of the triangle
$=\frac{1}{2}|\text{x}_1(\text{y}_2 -\text{y}_3) +\text{x}_2(\text{y}_3 -\text{y}_1) + \text{x}_3 (\text{y}_1 -\text{y}_2)|$
$=\frac{1}{2}|5(0-4)+8(4-0)+8(0)|$
$=\frac{1}{2}|-20+32+0$
$=\frac{1}{2}\times12$
$= 6 \text{sq. units}$

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MCQ 261 Mark

If A(-6, 7) and B(-1, -5) are two given points then the distance 2AB is:

A
13
✓
26
C
169
D
238

Answer

Correct option: B.

A(-6, 7) and B(-1, -5) are two given points.
Using the distance formula, we get
$2\text{AB}=\sqrt{(6-1)^2+(-7-5)^2}$
$=2\sqrt{(5)^2+(-12)^2}$
$=2\sqrt{25+144}$
$=2\sqrt{169}$
$=2\times13$
$=26$

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MCQ 271 Mark

The distance of the point (-3, 4) from x-axis is:

A
3
B
-3
✓
4
D
5

Answer

Correct option: C.

The distance of the point P(-3, 4) from the x-axis
= Y-coordinate of the point
= 4 units

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MCQ 281 Mark

If A(1, 3), B(-1, 2), C(2, 5) and D(x, 4) are the vertices of a || gm ABCD then the value of x is:

A
$3$
✓
$4$
C
$0$
D
$\frac{3}{2}$

Answer

Correct option: B.

$4$

Since ABCD is a ||gm, the diagonals bisect eachother.
So, Mis the mid-point of BD as well as AC.
$\frac{1+2}{2}=\frac{\text{x}-1}{2}$
$\Rightarrow 1+2 = \text{x}-1$
$\Rightarrow \text{x}=4$

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MCQ 291 Mark

The mid-point of segment AB is P(0, 4). If the coordinates of B are (-2, 3), then the coordinates of A are:

✓
(2, 5)
B
(-2, -5)
C
(2, 9)
D
(-2, 11)

Answer

Correct option: A.

(2, 5)

Let the mid-point of A be (x, y).P(0, 4) is given to be mid-point AB.
Using the mid-point formula, we get
$(0,4)=\Big(\frac{-2+\text{x}}{2},\frac{3+\text{y}}{2}\Big)$
$\Rightarrow\ 0=\frac{-2+\text{x}}{2}$ and $4=\frac{3+\text{y}}{2}$
⇒ -2 + x = 0 and 3 + y = 8
⇒ x = 2 and y = 5
So, the coordinates of A are (2, 5).

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MCQ 301 Mark

In what ratio does the y-axis divide the join of P(-4, 2) and Q(8, 3)?

A
3 : 1
B
1 : 3
C
2 : 1
✓
1 : 2

Answer

Correct option: D.

1 : 2

let the y-axis cut AB at the point P(0, y) in the ratio k : 1.Then, using section formula, we get
$\frac{8\text{k}-4}{\text{k}+1}=0$
$\Rightarrow\ 8\text{k}-4=0$
$\Rightarrow\ \text{k}=\frac{1}{2}$
So, the required ratio is $\frac{1}{2}:1,$ that is 1 : 2.

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MCQ 311 Mark

The perimeter of the triangle with vertices (0, 4), (0, 0) and (3, 0) is:

A
$(7+\sqrt{5})$
B
$5$
C
$10$
✓
$12$

Answer

Correct option: D.

$12$

AO = 4 units
BO = 3 units
Using Distance formula, we get:
$\text{AB}=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5\text{ units}$
So, the perimeter of the tringle
= AB + AO + BO
= 5 + 4 + 3
= 12 units

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MCQ 321 Mark

The distance of P(3, 4) from the x-axis is:

A
3 units
✓
4 units
C
5 units
D
1 units

Answer

Correct option: B.

4 units

The distance of the point P(3, 4) is given by the y-coordinate of the points P.
So, the distance = 4 units.

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MCQ 331 Mark

If P(-1, 1) is the mid-point of the line segment joining A(-3, b) and B(1, b + 4) then b = ?

A
1
✓
-1
C
2
D
0

Answer

Correct option: B.

-1

Given that P is the mid-point of AB.Using mid-point formula, we get
$\text{P(-1,1)}=\Big(\frac{-3+1}{2},\frac{\text{b}+\text{b}+4}{2}\Big)$
$\Rightarrow\ \frac{\text{b}+\text{b}+4}{2}=1$
$\Rightarrow 2\text{b} + 4 = 2$
$\Rightarrow 2\text{b} = -2$
$\Rightarrow\text{ b} = -1$

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MCQ 341 Mark

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonal is:

✓
5
B
4
C
3
D
25

Answer

Correct option: A.

The length of the diagonal BD $=\sqrt{(4-0)^2+(0-3)^2}$
$=\sqrt{16+9}$
$=5\text{ units}$

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