MCQ - MATHS STD 11 Science Questions - Vidyadip

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MCQ 11 Mark

Three vertices of a parallelogram taken in order are (-1, -6), (2, -5) and (7, 2). The fourth vertex is:

A
(1, 4)
✓
(4, 1)
C
(1, 1)
D
(4, 4)

Answer

Correct option: B.

(4, 1)

Let A(-1, -6), B(2, -5) and C(7, 2) be the given vertex. Let D(h, k) be the fourth vertex.
The midpoints of AC and BD are (3, -2) and $\Big(\frac{2+\text{h}}{2},\frac{-5+\text{k}}{2}\Big)$ respectively.
We know that the diagonals of a parallelogram bisect each other.
$\therefore3=\frac{2+\text{h}}{2}$ and $-2=\frac{-5+\text{k}}{2}$
$\Rightarrow\text{h}=4$ and $\text{k}=1$

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MCQ 21 Mark

The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1:

A
Lies in the III quadrant.
B
Lies in the II quadrant.
C
Lies in the I quadrant.
✓
Cannot be found.

Answer

Correct option: D.

Cannot be found.

The point which divides the join of (1, 2) and (3, 4) externally in the ratio 1 : 1 is
$\Big(\frac{1\times3-1\times1}{1-1},\frac{1\times4-1\times2}{1-1}\Big)$
which is not defined.
Therefore, it is not possible to externally divide the line joining two points in the ratio 1 : 1

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MCQ 31 Mark

L is a variable line such that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero. The line L will always pass through:

✓
(1, 1)
B
(2, 1)
C
(1, 2)
D
none of these.

Answer

Correct option: A.

(1, 1)

(1,1)

Solution:
Let ax + by + c = 0 be the variable line. It is given that the algebraic sum of the distances of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero
$\therefore \ \frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}=0$
$\Rightarrow3\text{a}+3\text{b}+3\text{c}=0$
$\Rightarrow\text{a}+\text{b}+\text{c}=0$
Substituting c = -a - b in ax + by + c = 0, we get:
$\text{ax}+\text{by}-\text{a}-\text{b}=0$
$\Rightarrow\text{a}(\text{x}-1)+\text{b}(\text{y}-1)=0$
$\Rightarrow(\text{x}-1)+\frac{\text{a}}{\text{b}}(\text{y}-1)=0$
This line is of the form $\text{L}_1+\lambda\text{L}_2=0$ which passes through the intersection of $L_1 =0$ and $L_2 =0$, i.e. x - 1 = 0 and y - 1 = 0.
⇒ x = 1, y = 1

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MCQ 41 Mark

The acute angle between the medians drawn from the acute angles of a right angled isosceles triangle is:

A
$\cos^{-1}\big(\frac{2}{3}\big)$
✓
$\cos^{-1}\big(\frac{3}{4}\big)$
C
$\cos^{-1}\big(\frac{4}{5}\big)$
D
$\cos^{-1}\big(\frac{5}{6}\big)$

Answer

Correct option: B.

$\cos^{-1}\big(\frac{3}{4}\big)$

$\cos^{-1}\big(\frac{3}{4}\big)$

Solution:
Let the coordinates of the right-angled isosceles triangle be O(0, 0), A(a, 0) and B(0, a).

Here, BD and AE are the medians drawn from the acute angles B and A, respectively.
$\therefore$ Slope of $BD = m_1$
$=\frac{0-\text{a}}{\frac{\text{a}}{2}-0}$
$=-\frac{1}{2}$
Let $\theta$ be the angle between BD and AE.
$\tan\theta=\Big|\frac{-2+\frac{1}{2}}{1+1}\Big|$
$=\frac{3}{4}$
$\Rightarrow\cos\theta=\frac{4}{\sqrt{3^2+4^2}}$
$\Rightarrow\cos\theta=\frac{4}{5}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{4}{5}\Big)$
Hence, the acute angle between the medians is $\cos^{-1}\Big(\frac{4}{5}\Big).$

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MCQ 51 Mark

The medians AD and BE of a triangle with vertices A(0, b), B(0, 0) and C(a, 0) are perpendicular to each other, if

A
$\text{a}=\frac{\text{b}}{2}$
B
$\text{b}=\frac{\text{a}}{2}$
C
$\text{ab}=1$
✓
$\text{a}=\pm\sqrt{2}\text{b}$

Answer

Correct option: D.

$\text{a}=\pm\sqrt{2}\text{b}$

The midpoints of BC and AC are $\text{D}\Big(\frac{\text{a}}{2},0\Big)$ and $\text{E}\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
Slope of $\text{AD}=\frac{0-\text{b}}{\frac{\text{a}}{2}-0}$
Slope of $\text{BE}=\frac{-\frac{\text{b}}{2}}{\frac{\text{-a}}{2}}$
It is given that the medians are perpendicular to each other.
$\frac{0-\text{b}}{\frac{\text{a}}{2}-0}\times\frac{-\frac{\text{b}}{2}}{-\frac{\text{a}}{2}}=-1$
$\Rightarrow\text{a}=\pm\sqrt{2}\text{b}$

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MCQ 61 Mark

The line segment joining the points (-3, -4) and (1, -2) is divided by y-axis in the ratio:

A
1 : 3
B
2 : 3
✓
3 : 1
D
3 : 2

Answer

Correct option: C.

3 : 1

Let the points (-3, -4) and (1, -2) be divided by y-axis at (0, t) in the ratio m : n.
$\therefore\Big(\frac{\text{m}-3\text{n}}{\text{m}+\text{n}},\frac{-2\text{m}-4\text{n}}{\text{m}+\text{n}}\Big)=(0,\text{t})$
$\Rightarrow0=\frac{\text{m}-3\text{n}}{\text{m}+\text{n}}$
$\Rightarrow\text{m}:\text{n}=3:1$

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MCQ 71 Mark

The number of real values of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent is:

✓
0
B
1
C
2
D
Infinite.

Answer

Correct option: A.

0

$\text{x} - 2\text{y} + 3 = 0 \ ...(\text{i})$
$\lambda\text{x} + 3\text{y} + 1 = 0 \ ...(\text{ii})$
$4\text{x} - \lambda\text{y} + 2 = 0 \ ...(\text{iii})$
It is given that (1), (2) and (3) are concurrent.
$\therefore\begin{vmatrix} 1&-2&3\\\lambda&3&1\\4&-\lambda&2\end{vmatrix}=0$
$\Rightarrow(6+\lambda)+2(2\lambda-4)+3(-\lambda^2-12)=0$
$\Rightarrow6+\lambda+4\lambda-8-3\lambda^2-36=0$
$\Rightarrow5\lambda-3\lambda^2-38=0$
$\Rightarrow3\lambda^2-5\lambda+38=0$
The discriminant of this equation is $25 - 4 \times 3 \times 38 = -431$
Hence, there is no real value of $\lambda$ for which the lines $\text{x} - 2\text{y} + 3 = 0, \lambda\text{x} + 3\text{y} + 1 = 0$ and $4\text{x} - \lambda\text{y} + 2 = 0$ are concurrent.

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MCQ 81 Mark

If the point (5, 2) bisects the intercept of a line between the axes, then its equation is:

A
5x + 2y = 20
✓
2x + 5y = 20
C
5x - 2y = 20
D
2x - 5y = 20

Answer

Correct option: B.

2x + 5y = 20

Let the equation of the line be $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
The coordinates of the intersection of this line with the coordinate axes are (a, 0) and (0, b).
The midpoint of (a, 0) and (0, b) is $\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)$
According to the question:
$\Big(\frac{\text{a}}{2},\frac{\text{b}}{2}\Big)=(5,2)$
$\Rightarrow\frac{\text{a}}{2}=5,\frac{\text{b}}{2}=2$
$\Rightarrow\text{a}=10,\text{b}=4$
The equation of the required line is given below:
$\frac{\text{x}}{10}+\frac{\text{y}}{4}=1$
$\Rightarrow2\text{x}+5\text{y}=20$

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MCQ 91 Mark

Distance between the lines 5x + 3y - 7 = 0 and 15x + 9y + 14 = 0 is:

A
$\frac{35}{\sqrt{34}}$
B
$\frac{1}{3\sqrt{34}}$
✓
$\frac{35}{3\sqrt{34}}$
D
$\frac{35}{2\sqrt{34}}$

Answer

Correct option: C.

$\frac{35}{3\sqrt{34}}$

The given lines can be written as
$5\text{x}+3\text{y}-7=0 \ ...(1)$
$5\text{x}+3\text{y}+\frac{14}{3}=0 \ ...(2)$
Let d be the distance between the lines 5x + 3y - 7 = 0 and 15x + 9y + 14 = 0
Then, $\text{d}=\Bigg|\frac{-7-\frac{14}{3}}{\sqrt{5^2+3^2}}\Bigg|$
$\Rightarrow\text{d}=\frac{35}{3\sqrt{34}}$

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MCQ 101 Mark

The area of a triangle with vertices at (-4, -1), (1, 2) and (4, -3) is:

✓
17
B
16
C
15
D
None of these.

Answer

Correct option: A.

17

Let A be the area of the triangle formed by the points (-4, -1), (1, 2) and (4, -3).
$\therefore\text{A}=\frac{1}{2}\big|\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\}\big|$
$\Rightarrow\text{A}=\frac{1}{2}\big|\{-4(2+3)+1(-3+1)+4(-1-2)\}\big|$
$\Rightarrow\text{A}=17$

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MCQ 111 Mark

The equation of the straight line which passes through the point (-4, 3) such that the portion of the line between the axes is divided internally by the point in the ratio 5 : 3 is:

✓
9x - 20y + 96 = 0
B
9x + 20y = 24
C
20x + 9y + 53 = 0
D
none of these.

Answer

Correct option: A.

9x - 20y + 96 = 0

Let the required line intersects the coordinate axis at (a, 0) and (0, b).

The point (−4, 3) divides the required line in the ratio 5 : 3
$\therefore \ -4=\frac{5\times0+3\times\text{a}}{5+3}$ and $3=\frac{5\times\text{b}+3\times0}{5+3}$
$\Rightarrow\text{a}=\frac{ -32}{3}$ and $\text{b}=\frac{ 24}{5}$
Hence, The equation of the required line is given below:
$\frac{\text{x}}{\frac{-32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$
$\Rightarrow\frac{-3\text{x}}{32}+\frac{5\text{y}}{24}=1$
$\Rightarrow-9\text{x}+20\text{y}=96$
$\Rightarrow9\text{x}-20\text{y}+96=0$

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MCQ 121 Mark

Two vertices of a triangle are (-2, -1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 square units, then the third vertex is:

A
(0, 5) or, (4, 1)
✓
(5, 0) or, (1, 4)
C
(5, 0) or, (4, 1)
D
(0, 5) or, (1, 4)

Answer

Correct option: B.

(5, 0) or, (1, 4)

Let (h, k) be the third vertex of the triangle.
It is given that the area of the triangle with vertices (h, k), (-2, -1) and (3, 2) is 4 square units.
$\frac{1}{2}\big|\text{h}(-1-2)-3(-1-\text{K})-2(2-\text{K})\big|=4$
$\Rightarrow3\text{h}-5\text{k}+1=\pm 8$
Taking positive sign, we get,
3h - 5k + 1 = 8
3h - 5k - 7 = 0 ...(1)
Taking negative sign, we get,
3h - 5k + 9 = 0 ...(2)
The vertex (h, k) lies on the line x + y = 5.
h + k - 5 = 0 ...(3)
On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex.
Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex.

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MCQ 131 Mark

If p be the length of the perpendicular from the origin on the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ then:

A
$\text{p}^2=\text{a}^2+\text{b}^2$
B
$\text{p}^2=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
✓
$\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
D
None of these.

Answer

Correct option: C.

$\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$

It is given that p is the length of the perpendicular from the origin on the line
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\frac{1}{\text{a}}\text{x}+\frac{1}{\text{b}}\text{y}-1=0$
$\therefore\text{p}=\begin{vmatrix}\frac{0+0+1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}}} \end{vmatrix}$
Squaring both sides,
$\Rightarrow\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$

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MCQ 141 Mark

The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the line 3x + 4y + 5 = 0 and 3x + 4y - 5 = 0 is:

A
1 : 2
✓
3 : 7
C
2 : 3
D
2 : 5

Answer

Correct option: B.

3 : 7

Here, in all equations the coefficient of x is same.
It means all the lines have same slope
So, all the lines are parallel.
Now, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y + 5 = 0 is given by
$\frac{|2-5|}{\sqrt{3^2+4^2}}$
$=\frac{3}{\sqrt{25}}=\frac{3}{5}$
Hence, the ratio is given by
$\frac{3}{5}:\frac{7}{5}$
$=3:7$

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MCQ 151 Mark

The inclination of the straight line passing through the point (-3, 6) and the mid-point of the line joining the point (4, -5) and (-2, 9) is:

A
$\frac{\pi}{4}$
B
$\frac{\pi}{6}$
C
$\frac{\pi}{3}$
✓
$\frac{3\pi}{4}$

Answer

Correct option: D.

$\frac{3\pi}{4}$

The midpoint of the line joining the points (4, -5) and (-2, 9) is (1, 2).
Let $\theta$ be the inclination of the straight line passing through the points (-3, 6) and (1, 2).
Then, $\tan\theta=\frac{ 2-6}{1+3}=-1$
$\Rightarrow\theta=\frac{3\pi}{4}$

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MCQ 161 Mark

The angle between the lines 2x - y + 3 = 0 and x + 2y + 3 = 0 is:

✓
90°
B
60°
C
45°
D
30°

Answer

Correct option: A.

90°

90°

Solution:
Let $m_1$ and $m_2$ be the slope of the lines 2x - y + 3 = 0 and x + 2y + 3 = 0, respectively.
Let $\theta$ be the angle between them.
Here, $m_1 = 2$ and $\text{m}_2=-\frac{1}{2}$
$\because\text{m}_1\text{m}_2=-1$
Therefore, the angle between the given lines is 90°.

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MCQ 171 Mark

A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are four points. If $\triangle\text{DBC} : \triangle\text{ABC}= 1 : 2,$ then x is equal to:

✓
$\frac{11}{8}$
B
$\frac{8}{11}$
C
3
D
None of these

Answer

Correct option: A.

$\frac{11}{8}$

The area of a triangle with vertices D(x, 3x), B(-3, 5) and C(4, -2) is given below:
Area of $\triangle\text{DBC}=\frac{1}{2}\{\text{x}(5+2)-3(-2-3\text{x})+4(3\text{x}-5)\}$
⇒ Area of $\triangle\text{DBC}=(14\text{x}-7)\text{sq units}$
Similarly, the area of a triangle with vertices A(6, 3), B(-3, 5) and C(4, -2) is given below:
$\triangle\text{ABC}=\frac{1}{2}\{6(5+2)-3(-2-3)+4(3-5)\}$
$\triangle\text{ABC}=\frac{49}{2}\text{sq units}$
Given:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$

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MCQ 181 Mark

The equations of the sides AB, BC and CA of $\triangle \text{ABC}$ are y - x = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B is:

A
x - 3y + 1 = 0
✓
x - 3y + 4 = 0
C
3x - y + 2 = 0
D
None of these.

Answer

Correct option: B.

x - 3y + 4 = 0

The equation of the sides AB, BC and CA of $\triangle \text{ABC}$ are y - x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively.
Solving the equations of AB and BC, i.e. y - x = 2 and x + 2y = 1, we get:
x = -1, y = 1
So, the coordinates of B are (-1, 1).
The altitude through B is perpendicular to AC.
$\therefore$ Slope of AC = -3
Thus, slope of the altitude through B is 13. Thus, slope of the altitude through B is $\frac{1}{3}.$
Equation of the required altitude is given below:
$\text{y}-1+\frac{1}{3}(\text{x}+1)$
$\Rightarrow\text{x}-3\text{y}+4=0$

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MCQ 191 Mark

If a + b + c = 0, then the family of lines 3ax + by + 2c = 0 pass through fixed point:

A
$\Big(2,\frac{2}{3}\Big)$
✓
$\Big(\frac{2}{3},2\Big).$
C
$\Big(-2,\frac{2}{3}\Big)$
D
None of these.

Answer

Correct option: B.

$\Big(\frac{2}{3},2\Big).$

$\Big(\frac{2}{3},2\Big).$

Solution:
Given:
a + b + c = 0
Substituting c = -a - b in 3ax + by + 2c = 0, we get:
$3\text{ax}+\text{by}-2\text{a}-2\text{b}=0$
$\Rightarrow\text{a}(3\text{x}-2)+\text{b}(\text{y}-2)=0$
$\Rightarrow(3\text{x}-2)+\frac{\text{b}}{\text{a}}(\text{y}-2)=0$
This line is of the form $\text{L}_1+\lambda\text{L}_2=0,$ which passes through the intersection of the lines $L_1$ and $L_2$ i.e. 3x - 2 = 0 and y - 2 = 0.
Solving 3x - 2 = 0 and y - 2 = 0, we get
$\text{x}=\frac{2}{3},\text{y}=2$
Hence, the required fixed point is $\Big(\frac{2}{3},2\Big).$

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MCQ 201 Mark

If the lines x + q = 0, y - 2 = 0 and 3x + 2y + 5 = 0 are concurrent, then the value of q will be:

A
1
B
2
✓
3
D
5

Answer

Correct option: C.

3

The lines x + q = 0, y - 2 = 0 and 3x + 2y + 5 = 0 are concurrent.
$\therefore\begin{vmatrix}1&0&\text{q}\\0&1&-2\\3&2&5 \end{vmatrix}=0$
$\Rightarrow1(5+4)-0+\text{q}(0-3)=0$
$\Rightarrow3\text{q}=9$
$\Rightarrow\text{q}=3$

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MCQ 211 Mark

A point equidistant from the line 4x + 3y + 10 = 0, 5x - 12y + 26 = 0 and 7x + 24y - 50 = 0 is:

A
(1, -1)
B
(1, 1)
✓
(0, 0)
D
(0, 1)

Answer

Correct option: C.

(0, 0)

Let the coordiantes of the point be (a, b)
Now, the distance of the point (a, b) from 4x + 3y + 10 = 0 is given by
$\Bigg|\frac{4\text{a}+3\text{b}+10}{\sqrt{4^2+3^2}}\Bigg|$
$=\Bigg|\frac{4\text{a}+3\text{b}+10}{5}\Bigg|$
Again, the distance of the point (a, b) from 5x - 12y + 26 = 0 is given by
$\Bigg|\frac{5\text{a}-12\text{b}+26}{\sqrt{5^2+(-12)^2}}\Bigg|$
$=\Bigg|\frac{5\text{a}-12\text{b}+26}{13}\Bigg|$
Again, the distance of the point (a, b) from 7x + 24y - 50 = 0 is is given by
$\Bigg|\frac{7\text{a}+24\text{b}-50}{\sqrt{7^2+(24)^2}}\Bigg|$
$=\Bigg|\frac{7\text{a}+24\text{b}-50}{25}\Bigg|$
Now,
$\Bigg|\frac{4\text{a}+3\text{b}+10}{5}\Bigg|=\Bigg|\frac{5\text{a}-12\text{b}+26}{13}\Bigg|=\Bigg|\frac{7\text{a}+24\text{b}-50}{25}\Bigg|$
Only a = 0 and b = 0 is satisfying the above equation
Hence, the correct answer is option (c).

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MCQ 221 Mark

Area of the triangle formed by the points ((a + 3)(a + 4), a + 3), ((a + 2)(a + 3), (a + 2)) and ((a + 1)(a + 2), (a + 1)) is:

A
$25 a^2$
B
$5 a^2$
C
$24 a^2$
✓
None of these.

Answer

Correct option: D.

None of these.

None of these.

Solution:
The given points are {(a + 3)(a + 4), a + 3), ((a + 2)(a + 3), (a + 2)) and ((a + 1)(a + 2), (a + 1)}
Let A be the area of the triangle formed by these points.
Then, $\text{A}=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)(\text{a}+2-\text{a}-1)+(\text{a}+2)(\text{a}+3)\$\text{a}+1-\text{a}-3)+(\text{a}+1)(\text{a}+2)(\text{a}+3-\text{a}-2)]$
$\Rightarrow\text{A}=\frac{1}{2}[(\text{a}+3)(\text{a}+4)-2(\text{a}+2)(\text{a}+3)+(\text{a}+1)(\text{a}+2)]$
$\Rightarrow\text{A}=\frac{1}{2}[\text{a}^2+7\text{a}+12-2\text{a}^2-10\text{a}-12+\text{a}^2+3\text{a}+2]$
$\Rightarrow\text{A}=1$

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MCQ 231 Mark

The equation of the line with slope $-\frac{3}{2}$ and which is concurrent with the lines 4x + 3y - 7 = 0 and 8x + 5y - 1 = 0 is:

A
3x + 2y - 63 = 0
✓
3x + 2y - 2 = 0
C
2y - 3x - 2 = 0
D
None of these.

Answer

Correct option: B.

3x + 2y - 2 = 0

Given:
4x + 3y - 7 = 0 ...(1)
8x + 5y - 1 = 0 ...(2)
The equation of the line with slope $-\frac{3}{2}$ is given below:
$\text{y}=-\frac{3}{2}\text{x}+\text{c}$
$\Rightarrow\frac{3}{2}\text{x}+\text{y}-\text{C}=0 \ ...(3)$
The lines (1), (2) and (3) are concurrent.
$\therefore\begin{vmatrix}4&3&-7\\8&5&-1\\\frac{3}{2}&1&\text{-c} \end{vmatrix}=0$
$\Rightarrow4(-5\text{c}+1)-3\Big(-8\text{c}+\frac{3}{2}\Big)-7\Big(8-\frac{15}{2}\Big)=0$
$\Rightarrow-20\text{c}+4+24\text{c}-\frac{9}{2}-56+\frac{105}{2}=0$
$\Rightarrow\frac{-40\text{c}+8+48\text{c}-9-112+105}{2}=0$
$\Rightarrow8\text{c}=8$
$\Rightarrow\text{c}=1$
On substituting c = 1 in $\text{y}=-\frac{3}{2}\text{x}+\text{c},$ we get:
$\text{y}=-\frac{3}{2}\text{x}+1,$
$\Rightarrow3\text{x}+2\text{y}-2=0$

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MCQ 241 Mark

If $p_1$ and $p_2$ are the lengths of the perpendiculars from the origin upon the lines $\text{x} \sec \theta + \text{y} \text{cosec}\theta = \text{a}$ and $\text{x} \cos \theta - \text{y} \sin \theta = \text{a} \cos 2 \theta$ respectively, then:

✓
$4\text{p}_1^2 + \text{p}_2^2 = \text{a}^2$
B
$\text{p}_1^2 + 4\text{p}_2^2 = \text{a}^2$
C
$\text{p}_1^2 + 4\text{p}_2^2 = \text{a}^2$
D
None of these.

Answer

Correct option: A.

$4\text{p}_1^2 + \text{p}_2^2 = \text{a}^2$

$4\text{p}_1^2 + \text{p}_2^2 = \text{a}^2$

Solution:
The given lines are
$\text{x} \sec \theta + \text{y} \text{cosec}\theta = \text{a} \ ...(1)$
$\text{x} \cos \theta - \text{y} \sin \theta = \text{a} \cos 2 \theta \ ...(2)$
$p_1$ and $p_2$ are the perpendiculars from the origin upon the lines (1) and (2), respectively.
$\text{p}_1=\Big|\frac{-\text{a}}{\sqrt{\sec^2\theta+\text{cosec}^2}\theta}\Big|$ and $\text{p}_2=\Big|\frac{-\text{a}\cos2\theta}{\sqrt{\cos^2\theta+\sin^2}\theta}\Big|$
$\Rightarrow\text{p}_1=\Big|\frac{-\text{a}\sin\theta\cos\theta}{\sqrt{\sin^2\theta+\cos^2}\theta}\Big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow\text{p}_1=\frac{1}{2}\big|-\text{a}\times2\sin\theta\cos\theta\big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow\text{p}_1=\frac{1}{2}\big|-\text{a}\times\sin2\theta\big|$ and $\text{p}_2=\big|-\text{a}\cos2\theta\big|$
$\Rightarrow4\text{p}_1^2+\text{p}_2^2=\text{a}^2(\sin^22\theta+\cos^22\theta)$
$=\text{a}^2$

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MCQ 251 Mark

The figure formed by the lines $\text{ax} \pm \text{by} \pm \text{c} = 0 $ is:

A
a rectangle.
B
a square.
✓
a rhombus.
D
None of these.

Answer

Correct option: C.

a rhombus.

The given lines can be written separately in the following manner:
ax + by + c = 0 ...(1)
ax + by - c = 0 ...(2)
ax - by - c = 0 ...(3)
ax - by - c = 0 ...(4)
Graph of the given lines is given below:

Clearly, $\text{AB}=\text{BC}=\text{CD}=\text{DA}=\sqrt{\frac{\text{a}^2}{\text{c}^2}+\frac{\text{b}^2}{\text{c}^2}}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{|\text{c}|}$
Thus, the region formed by the given lines is ABCD, which is a rhombus.

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MCQ 261 Mark

The line segment joining the points (1, 2) and (-2, 1) is divided by the line 3x + 4y = 7 in the ratio:

A
3 : 4
B
4 : 3
C
9 : 4
✓
4 : 9

Answer

Correct option: D.

4 : 9

Let the line segment joining the points (1, 2) and (−2, 1) be divided by the line 3x + 4y = 7 in the ratio m:n.
Then, the coordinates of this point will be $\Big(\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}},\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}\Big)$ that lie on the line.
$3\text{x}+4\text{y}=7$
$3\times\frac{-2\text{m}+\text{n}}{\text{m}+\text{n}}+4\times\frac{\text{m}+2\text{n}}{\text{m}+\text{n}}=7$
$\Rightarrow-2\text{m}+11\text{n}=7\text{m}+7\text{n}$
$\Rightarrow-9\text{m}=-4\text{n}$
$\Rightarrow\text{m}:\text{n}=4:9$

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MCQ 271 Mark

A line passes through the point (2, 2) and is perpendicular to the line 3x + y = 3. Its y-intercept is:

A
$\frac{1}{3}$
B
$\frac{2}{3}$
C
1
✓
$\frac{4}{3}$

Answer

Correct option: D.

$\frac{4}{3}$

The equation of the line perpendicular to 3x + y = 3 is given below:
$\text{x}-3\text{y}+\lambda=0$
This line passes through (2, 2).
$2-6+\lambda=0$
$\Rightarrow\lambda=4$
So, the equation of the line will be
x - 3y + 4 = 0
$\Rightarrow\text{y}=\frac{1}{3}\text{x}+\frac{4}{3}$
Hence, the y-intercept is $\frac{4}{3}.$

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MCQ 281 Mark

If the lines ax + 12y + 1 = 0, bx + 13y + 1 = 0 and cx + 14y + 1 = 0 are concurrent, then a, b, c are in:

A
H.P.
B
G.P.
✓
A.P.
D
None of these.

Answer

Correct option: C.

A.P.

The given lines are
ax + 12y + 1 = 0 ...(1)
bx + 13y + 1 = 0 ...(2)
cx + 14y + 1 = 0 ...(3)
It is given that (1), (2) and (3) are concurrent.
$\begin{vmatrix} \text{a}&12&1\\\text{b}&13&1\\\text{c}&14&1\end{vmatrix}=0$
$\Rightarrow\text{a}(13-14)-12(\text{b}-\text{c})+14\text{b}-13\text{c}=0$
$\Rightarrow-\text{a}-12\text{b}+12\text{c}+14\text{b}-13\text{c}=0$
$\Rightarrow-\text{a}+2\text{b}-\text{c}=0$
$\Rightarrow2\text{b}=\text{a}+\text{c}$
Hence, a, b and c are in AP.

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MCQ 291 Mark

The vertices of a triangle are (6, 0), (0, 6) and (6, 6). The distance between its circumcentre and centroid is:

A
$2\sqrt{2}$
B
$2$
✓
$\sqrt{2}$
D
$1$

Answer

Correct option: C.

$\sqrt{2}$

Let A(0, 6), B(6, 0) and C(6, 6) be the vertices of the given triangle.

Centroid of $\triangle\text{ABC}=\Big(\frac{0+6+6}{3},\frac{6+0+6}{3}\Big)$
$=(4,4)$
Coordinates of $\text{N}=\Big(\frac{6+6}{2},\frac{6+0}{2}\Big)$
$=(6,3)$
Coordinates of $\text{P}=\Big(\frac{0+6}{2},\frac{6+6}{2}\Big)$
$=(3,6)$
Equation of MN is y = 3
Equation of MP is x = 3
As, we know that circumcentre of a triangle is the intersection of the perpendicular bisectors of any two sides.
Therefore, coordinates of circumcentre is (3, 3)
Thus, the coordinates of the circumcentre are (3, 3) and the centroid of the triangle is (4, 4).
Let d be the distance between the circumcentre and the centroid.
$\therefore\text{d}\sqrt{(4-3)^2+(4-3)^2}=\sqrt{2}$

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MCQ 301 Mark

The equation of the line passing through (1, 5) and perpendicular to the line 3x - 5y + 7 = 0 is:

✓
5x + 3y - 20 = 0
B
3x - 5y + 7 = 0
C
3x - 5y + 6 = 0
D
5x + 3y + 7 = 0

Answer

Correct option: A.

5x + 3y - 20 = 0

A line perpendicular to 3x - 5y + 7 = 0 is given by
$5\text{x}+3\text{y}+\lambda=0$
This line passes through (1, 5)
$5+15+\lambda=0$
$\Rightarrow\lambda=-20$
Therefore, the equation of the required line is 5x + 3y - 20 = 0.

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MCQ 311 Mark

The reflection of the point (4, -13) about the line 5x + y + 6 = 0 is:

✓
(-1, -14)
B
(3, 4)
C
(0, 0)
D
(1, 2)

Answer

Correct option: A.

(-1, -14)

Let the reflection point be A(h, k)
Now, the mid point of line joining (h, k) and (4, -13) will lie on the line 5x + y + 6 = 0
$\therefore5\Big(\frac{\text{h}+4}{2}\Big)+\frac{\text{k}-13}{2}+6=0$
$\Rightarrow5\text{h}+20+\text{k}-13+12=0$
$\Rightarrow5\text{h}+\text{k}+19=0 \ ...(1)$
Now, the slope of the line joining points (h, k) and (4, -13) are perpendicular to the line 5x + y + 6 = 0.
slope of the line = -5
slope of line joining by points (h, k) and (4, -13)
$\frac{\text{k}+13}{\text{h}-4}$
$\therefore\frac{\text{k}+13}{\text{h}-4}(-5)=-1$
$\Rightarrow5\text{k}-\text{h}+60=0 \ ...(2)$
Solving (1) and (2), we get
h = -1 and k = -14

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MCQ 321 Mark

The value of $\lambda$ for which the lines $3\text{x} + 4\text{y} = 5, 5\text{x} + 4\text{y} =4$ and $\lambda\text{x} + 4\text{y} = 6$ meet at a point is:

A
2
✓
1
C
4
D
3

Answer

Correct option: B.

1

It is given that the lines $3\text{x} + 4\text{y} = 5, 5\text{x} + 4\text{y} =4$ and $\lambda\text{x} + 4\text{y} = 6$ meet at a point. In other words, the given lines are concurrent.
$\begin{vmatrix}3&4&-5\\5&4&-4\\\lambda&4&-6 \end{vmatrix}=0$
$\Rightarrow3(-24+16)-4(-30+4\lambda)-5(20-4\lambda)=0$
$\Rightarrow-24+120-16\lambda-100+20\lambda=0$
$\Rightarrow4\lambda=4$
$\Rightarrow\lambda=1$

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MCQ 331 Mark

The centroid of a triangle is (2, 7) and two of its vertices are (4, 8) and (-2, 6). The third vertex is:

A
(0, 0)
✓
(4, 7)
C
(7, 4)
D
(7, 7)

Answer

Correct option: B.

(4, 7)

Let A(4, 8) and B(-2, 6) be the given vertex. Let C(h, k) be the third vertex.
The centroid of $\triangle\text{ABC}$ is $\Big(\frac{4-2+\text{h}}{3},\frac{8+6+\text{k}}{3}\Big)$
It is given that the centroid of triangle ABC is (2, 7).
$\therefore\frac{4-2+\text{h}}{3}=2,\frac{8+6+\text{k}}{3}=7$
$\Rightarrow\text{h}=4,\text{h}=7$
Thus, the third vertex is (4, 7).

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MCQ 341 Mark

The distance between the orthocentre and circumcentre of the triangle with vertices (1, 2), (2, 1) and $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ is:

✓
0
B
$\sqrt{2}$
C
$3+\sqrt{3}$
D
none of these.

Answer

Correct option: A.

0

Let A(1, 2), B(2, 1) and C $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ be the given points.
$\therefore \ \text{AB}=\sqrt{(2-1)^2+(1-2)^2}$
$=\sqrt{2}$
$\text{BC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2}$
$=\sqrt{2}$
$\text{AC}=\sqrt{\Big(\frac{3+\sqrt{3}}{2}-1\Big)^2,\Big(\frac{3+\sqrt{3}}{2}-2\Big)^2}$
$=\sqrt{2}$
Thus, ABC is an equilateral triangle.
We know that the orthocentre and the circumcentre of an equilateral triangle are same.
So, the distance between the the orthocentre and the circumcentre of the triangle with vertices (1, 2), (2, 1) and $\Big(\frac{3+\sqrt{3}}{2},\frac{3+\sqrt{3}}{2}\Big)$ is 0.

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MCQ 351 Mark

The coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0 are:

A
(-6, 5)
✓
(5, 6)
C
(-5, 6)
D
(6, 5)

Answer

Correct option: B.

(5, 6)

Let the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y - 11 = 0 be (x, y)
Now, the slope of the line x + y - 11 = 0 is -1
So, the slope of the perpendicular = 1
The equation of the perpendicular is given by
y - 3 = 1(x - 2)
⇒ x - y + 1 = 0
Solving x + y - 11 = 0 and x - y + 1 = 0, we get
x = 5 and y = 6

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MCQ - MATHS STD 11 Science Questions - Vidyadip