# 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 _______

(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 __________

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 ____________

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 ____________

-20x + 9y + 96 = 0

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

cannot be found

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

4

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

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 __________

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 __________

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 =  __________

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.

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).

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.

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.

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).

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.

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.

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.

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.

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.

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.