CBSE Solutions for Class 11 Maths

Select CBSE Solutions for class 10 Subject & Chapters Wise :

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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Answer :

(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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Answer :

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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Answer :

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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Answer :

-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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Answer :

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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Answer :

4

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

 

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Answer :

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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Answer :

0

The equations of the sides ABBC 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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Answer :

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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Answer :

a2

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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Answer :

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 (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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Answer :

Let the given points be (3, 1) and (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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Answer :

Let the given points be (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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Answer :

Let the given points be (2, 7), (-2, 8), P (8, 12) and (x, 24).

Slope of AB = m1 = 8-7/2+2 =1/4

Slope of PQ = m= 24-12/x-8= 12/x-8

It is given that the line joining A (2, 7) and (-2, 8) and the line joining P (8, 12) and (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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Answer :

Let the given points be A (3, −1) and (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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Answer :

Let (0, −1), (6, 0), C (5, 3) and (−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 y − 11 = 0.

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Answer :

Let the coordinates of the foot of the perpendicular from the point (2, 3) on the line y − 11 = 0 be (xy)
Now, the slope of the line 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 y − 11 = 0 and − y + 1 = 0, we get
x = 5 and y = 6
coordinates are (5, 6)

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

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Answer :

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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Answer :

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

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

∴ 9 = 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 = passes through (2, 9).

∴ 2 = 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 = 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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Answer :

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

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

∴ 3 = 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 = passes through (2, 3).

∴ 2 = 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 = 2 and y = 3.

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The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.

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