CBSE Solutions for Class 11 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

Sets Relation and Function Trignometric Functions Mathematical Induction Complex Numbers and Quardratic Equations Linear Equations Permutation and Combination Binomial Theorem Sequence and Series Straight Lines Conic Section Introduction to 3D Geometry Limits and Derivatives Mathematical Reasoning Statistics Probability
Area of the triangle formed by the points [(x+3)(x+4), (x+3)], [(x+2)(x+3), (x+2)] and [(x+1)(x+2),(x+1)] is 24x2

 

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જવાબ : False


The angle between the lines x – 2y = 3 and y – 2x = 3 is tan-1 (2/3)

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જવાબ : False


Equation of the line passing through (0, 0) and slope n is y = nx

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જવાબ : True


The slope of the line px + qy + c = 0 is -p/q

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જવાબ : True


The slope of the line px + qy + c = 0 is -p/q

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જવાબ : True


Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are  if a1/a2 = b1/b2 = c1/c2 , then lines are coincedent

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જવાબ : True


The equation of the line which cuts off equal and positive intercepts from the axes and passes through the point (a, b) is x + y = a

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જવાબ : False


if a line if its slope is negative, θ is an obtuse angle

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જવાબ : True


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

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જવાબ : (1, 1)


In triangle ABC, rt. angled at B, The acute angle between the medians drawn from the A & C of a right angled isosceles triangle is __________

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જવાબ : cos−1(4/5)


The distance between the orthocentre and circumcentre of the triangle with vertices [1, (3+√3)/2], [(3+√3)/2, 1] and [2, 2] is ____________

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જવાબ : 0


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

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જવાબ : -20x + 9y + 96 = 0


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

 

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જવાબ : cannot be found


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

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જવાબ : 4


If the lines ax + 11y + 1 = 0, bx + 12y + 1 = 0 and cx + 13y + 1 = 0 are concurrent, then a, b, c are in __________

 

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જવાબ : A.P.


The number of real values of λ for which the lines x − 2y + 3 = 0, λx + 3y + 1 = 0 and 4x − λy + 2 = 0 are concurrent is __________

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જવાબ : 0


The equations of the sides AB, BC and CA of ∆ ABC are  x − y = -2, -x - 2y = -1 and -3x - y = 5 respectively. The equation of the altitude through B is __________

 

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જવાબ : x + 4 = 3y


If p1 and p2 are the lengths of the perpendiculars from the origin upon the lines x sec θ + y cosec θ = a and x cos θ − y sin θ = a cos 2 θ respectively, then 4p12 + p22 =  __________

 

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જવાબ : a2


Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are parallel if a1/a2 = __________

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જવાબ : b1/b2 ≠ c1/c2


If p + q + r = 0, then the family of lines 3px + bq + 2r = 0 pass through fixed point __________

 

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જવાબ : (2/3, 2)


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

 

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જવાબ : y axis


The area of a triangle is 17 than vertices are (−4, −1), (1, 2) and __________

 

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જવાબ : (4, −3)


The line segment joining the points (1, 2) and (−2, 1) is divided in the ratio 4 : 9 by __________

 

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જવાબ : 3x + 4y = 7


If the point (5, 2) divides the intercept of 3x + 4y = 7 between the axes, then ratio is __________

 

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જવાબ : 1:1


A (6, 3), B (−3, 5), C (4, −2) and D (11/8, 33/8) are four points. Then ∆ DBC : ∆ ABC =  __________

 

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જવાબ : 1 : 2


If α  be the length of the perpendicular from the origin on the line x/a + y/b = 1, then 1/α2 =  __________

 

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જવાબ : 1/a2 + 1/b2


The point of intersection where the 5x + 3y − 20 = 0 perpendicular to the line 3x − 5y + 7 = 0 is __________

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જવાબ : (1, 5)


The equation of straight line passing through the point (3, 10) and parallel to the line y – 2 = 3 × (x – 1) is y + 2 = 3 × (x + 1)

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જવાબ : False


The locus of a point, whose abscissa and ordinate are always equal is  x = y

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જવાબ : True


The sum of squares of the distances of a moving point from two fixed points (p, 0) and (-p, 0) is equal to 2q² then the equation of its locus is x² + y² = q² + p²

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જવાબ : False


The figure formed by the lines ax ± by = ± c is __________

 

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જવાબ : a rhombus


Two vertices of a triangle are (−2, −1) and (3, 2) and third vertex lies on the line (2, 3). If the area of the triangle is 4 square units, then the eqation of the line on 3rd vertex is __________

 

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જવાબ : x + y = 5


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 __________

 

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જવાબ : 3 π/4


If the distance is 35/3√34, between the lines 5x + 3y − 7 = 0 and 2nd is __________

 

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જવાબ : 15x + 9y + 14 = 0


The angle between the lines is 90°, 1st line is 2x − y + 3 = 0 and 2nd is  __________

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જવાબ : x + 2y + 3 = 0


The value of λ for which the lines 3x + 4y = 5, 5x + 4y = 4 and x + λy = 6 meet at a point is _________

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જવાબ : 4


Three vertices of a parallelogram taken in order are (−1, −6), (2, −5) and (4, 1). The fourth vertex is __________

 

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જવાબ : (7, 2)


If vertices are (4, 8), (4, 7)and (−2, 6).  The centroid of a triangle is __________

 

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જવાબ : (2, 7)


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

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જવાબ : 5


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= __________

 

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જવાબ : ±√2b


The equation of the line with slope −3/2 and which is concurrent with the lines 4x + 3y − 7 = 0 and 8x + 5y − 1 = 0 is __________

 

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જવાબ : 3x + 2y − 2 = 0


The vertices of a triangle are (6, 0), (0, 6) and (6, 6). The centroid is __________

 

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જવાબ : (4, 4)


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

 

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જવાબ : (0, 0)


3x + 4y + 2 = 0 divides the distance between the line 3x + 4y + 5 = 0 and 3x + 4y − 5 = 0 in ratio  __________

 

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જવાબ : 3: 7


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

 

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જવાબ : (5, 6)


The reflection of the point (−1, −14) about the line 5x + y + 6 = 0 is __________

 

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જવાબ : (4, −13).


The locus of a point, whose abscissa and ordinate are always equal is __________

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જવાબ : p – q = 0


In a ΔABC, if B is the point (7, – 2) and equations of the median through A and C are respectively x + y = 5 and x = 4, then A is __________

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જવાબ : (1, 2)


The length of the perpendicular from the origin to a line is 6 and the line makes an angle of 150 degrees with the positive direction of the y-axis. Then the equation of line is __________

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જવાબ : √3x + y = 12


If two vertices of a triangle are (3, -2) and (-5, -3) and its orthocenter is (-6, 1) then its third vertex is __________

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જવાબ : (-2, 3)


Find the equation of the line parallel to y-axis and having intercept − 2 on x-axis.

 

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જવાબ : The equation of a line parallel to the y-axis is x = k.

It is given that x = k has intercept −2 on the x-axis. This means that the line x = k  passes through (0, −2).

∴ −2 = k
⇒⇒ k = −2

Hence, the equation of the required line is x = −2.


Find the equation of the line parallel to x-axis and having intercept − 9 on y-axis.

 

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જવાબ : The equation of a line parallel to the x-axis is y = k.

It is given that y = k has intercept −9 on the y-axis. This means that the line y = k  passes through (0, −2).

∴ −9 = k
⇒⇒ k = −9

Hence, the equation of the required line is y = −9.


In the triangle ABC with vertices A (2, 3), B (4, −1) and C (1, 2), find the equation of the altitude from the vertex A.

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જવાબ : Equation of side BC:

y+1=2+1/1-4 (x-4)

⇒x+y-3=0

The equation of the altitude that is perpendicular to x+y-3=0 is x-y+λ=0.

Line x-y+λ=0 passes through (2, 3).

∴2-3+λ=0⇒λ=1

Thus, the equation of the altitude from the vertex A (2, 3) is x-y+1=0.


Find the angle between X-axis and the line joining the points (3, 1) and (4, 2).

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જવાબ : Let the given points be A (3, 1) and B (4, 2).

∴ Slope of AB = 2-1/4-3 = 1

Let θ be the angle between the x-axis and AB.

∴ tanθ=1

⇒θ=tan-1(1) = π/4


Find the value of x for which the points (2x, −2), (4, 2) and (8, 10) are collinear.

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જવાબ : Let the given points be A (2x, −2), B (4, 2) and C (8, 10).
Slope of AB = 2+2/4-2x = 2/2-x
Slope of BC = 10-2/8-4 = 4/2 = 2
It is given that the points (2x, −2), (4, 2) and (8, 10) are collinear.
∴ Slope of AB  = Slope of BC
⇒2/2-x=2

⇒1=2-x

⇒x=1

Hence, the value of x is 1.


Line through the points (2, 7) and (-2, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.

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જવાબ : Let the given points be A (2, 7), B (-2, 8), P (8, 12) and Q (x, 24).

Slope of AB = m1 = 8-7/2+2 =1/4

Slope of PQ = m2 = 24-12/x-8= 12/x-8

It is given that the line joining A (2, 7) and B (-2, 8) and the line joining P (8, 12) and Q (x, 24) are perpendicular.

∴ m1m2=-1

⇒ ¼ × 12/x-8 =-1

⇒x-8 = -3

⇒x=5

Hence, the value of x is 5.


Find the angle between the X-axis and the line joining the points (3, 1) and (4, 2).

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જવાબ : Let the given points be A (3, −1) and B (4, −2).

∴ Slope of AB = 2-1/4-3= 1

Let θ be the angle between the x-axis and AB.

∴ tanθ=1

⇒θ = tan-1(1) = 450


Without using the distance formula, show that points (0, −1), (6, 0), (5, 3) and (−1, 2) are the vertices of a parallelogram.

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જવાબ : Let A (0, −1), B (6, 0), C (5, 3) and D (−1, 2) be the given points.

Now, slope of AB= 0+1/0+2 = 1/2

Slope of BC= 3-0/5-4 =3

Slope of CD= 2-3/1-3 =1/2

Slope of DA= -1-2/2-1=3

Clearly, we have,
Slope of AB = Slope of CD
Slope of BC = Slope of DA

As the slopes of opposite sides are equal,
Therefore, both pair of opposite sides are parallel.
Hence, the given points are the vertices of a parallelogram.


Find the coordinates of the foot of the perpendicular from the point (2, 3) on the line x + y − 11 = 0.

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જવાબ : 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
coordinates are (5, 6)


Find the angle between the lines x = a and y = b.

 

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જવાબ : The given lines can be written as

x = a           ... (1)

y = b       ... (2)

Lines (1) and (2) are parallel to the y-axis and x-axis, respectively. Thus, they intersect at right angle, i.e. at 90°.


Find the angle between the lines x = a and y + c = 0.

 

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જવાબ : The given lines can be written as

x = a           ... (1)

y=−c      ... (2)

Lines (1) and (2) are parallel to the y-axis and x-axis, respectively. Thus, they intersect at right angle, i.e. at 90°.


Prove that the points (4, −2), (0, 4), (4, 6) and (8, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.

 

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જવાબ : Let A(4, −2), B(0, 4), C(4, 6) and D(8, 0) be the vertices.

Slope of AB = -3/2

Slope of BC = 1/2

Slope of CD = -3/2

Slope of DA = 1/2

Thus, AB is parallel to CD and BC is parallel to DA.

Therefore, the given points are the vertices of a parallelogram.


Prove that the lines 4x − 6y + 2 = 0, 2x + 2y = 6, 4x − 6y = 4  and 2x + 2y = 8 form a parallelogram.

 

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જવાબ : The given lines can be written as

y=2/3x+1/3            ... (1)

y=-x+3               ... (2)

y=2/3x-2/3            ... (3)
 
y=-x+4               ... (4)

The slope of lines (1) and (3) is 23 and that of lines (2) and (4) is −1.

Thus, lines (1) and (3), and (2) and (4) are two pair of parallel lines.

If both pair of opposite sides are parallel then ,we can say that it is a parallelogram.

Hence, the given lines form a parallelogram.


Find the equations of the altitudes of a ∆ ABC whose vertices are A (2, 8), B (−6, 4) and C (−10, −6).

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જવાબ : The vertices of ∆ABC are A (2, 8), B (−6, 4) and C (−10, −6).

Slope of AB = 1/2

Slope of BC = 5/2

Slope of CA = 7/6

Thus, we have:

Slope of CF = -2

Slope of AD = -25

Slope of BE = -67

Hence,

Equation of CF is : 2x+y+26=0

Equation of AD is : 2x+5y-44=0

Equation of BE is : 6x+7y+8=0


Find the equation of a line passing through the point (2, 4) and parallel to the line 3x − 4y + 5 = 0.

 

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જવાબ : The equation of the line parallel to 3x − 4y + 5 = 0 is 3x-4y+λ=0, where λ is a constant.

It passes through (2, 4).

∴6-16+λ=0⇒λ=10

Hence, the required line is 3x − 4y + 10 = 0


Find the equation of a line passing through the point (2, 2) and parallel to the line 3x − 4y + 5 = 0.

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જવાબ : The equation of the line parallel to 3x − 4y + 5 = 0 is 3x-4y+λ=0, where λ is a constant.

It passes through (2, 2).

∴6-8+λ=0

⇒λ=2

Hence, the required line is 3x − 4y + 2 = 0


Put the equation x/a + y/b =1 to the slope intercept form and find its slope.

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જવાબ : The given equation is x/a+ y/b=1

     bx + ay = ab

⇒ay = -bx+ab

⇒y=-b/ax + b

This is the slope intercept form of the given line.

∴ Slope = -b/a


Find the equations of the straight lines which pass through (2, 9) and are respectively parallel and perpendicular to the x-axis.

 

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જવાબ : The equation of the line parallel to the x-axis is y = b.

It is given that y = b passes through (2, 9).

∴ 9 = b
⇒ b = 9

Thus, the equation of the line parallel to the x-axis and passing through (2, 9) is y = 9.

Similarly, the equation of the line perpendicular to the x-axis is x = a.

It is given that x = a passes through (2, 9).

∴ 2 = a
⇒ a = 2

Thus, the equation of the line perpendicular to the x-axis and passing through (2, 9) is x = 2.

Hence, the required lines are x = 2 and y = 9.


Find the equations of the straight lines which pass through (2, 3) and are respectively parallel and perpendicular to the x-axis.

 

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જવાબ : The equation of the line parallel to the x-axis is y = b.

It is given that y = b passes through (2, 3).

∴ 3 = b
⇒ b = 3

Thus, the equation of the line parallel to the x-axis and passing through (2, 3) is y = 3.

Similarly, the equation of the line perpendicular to the x-axis is x = a.

It is given that x = a passes through (2, 3).

∴ 2 = a
⇒ a = 2

Thus, the equation of the line perpendicular to the x-axis and passing through (2, 3) is x = 2.

Hence, the required lines are x = 2 and y = 3.


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