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CBSE Solutions for Class 11 English

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

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 abc 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 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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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 + 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 12 =  __________

 

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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 4+ 3y + 10 = 0, 5x − 12y + 26 = 0 and 7x+ 24y − 50 = 0 is __________

 

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જવાબ : (0, 0)


3x + 4+ 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 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 k.

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

∴ −2 = k
⇒⇒ = −2

Hence, the equation of the required line is = −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 k.

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

∴ −9 = k
⇒⇒ = −9

Hence, the equation of the required line is = −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 (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 (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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જવાબ : 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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જવાબ : 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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જવાબ : 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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જવાબ : 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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જવાબ : 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)


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 = 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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જવાબ : 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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