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 24*x*^{2}

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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 a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are if a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2 }, 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

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 : -20*x* + 9*y* + 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 3*x* + *y* = 3. Its *x*-intercept is__________

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જવાબ : 4

If the lines *ax* + 11*y* + 1 = 0, *bx* + 12*y* + 1 = 0 and *cx* + 13*y* + 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* − 2*y* + 3 = 0, λ*x* + 3*y* + 1 = 0 and 4*x* − λ*y* + 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* - 2*y* = -1 and -3*x* - *y* = 5 respectively. The equation of the altitude through *B* is __________

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જવાબ : *x* + 4 = 3y

If *p*_{1} and *p*_{2} 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 4*p*_{1}^{2} + *p*_{2}^{2} = __________

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જવાબ : *a*^{2}

Two lines a_{1}x + b_{1}y + c_{1} = 0 and a_{2}x + b_{2}y + c_{2} = 0 are parallel if a_{1}/a_{2 = }__________

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જવાબ : b_{1}/b_{2} ≠ c_{1}/c_{2}

If *p* + *q *+ *r* = 0, then the family of lines 3*px* + *bq* + 2*r* = 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 : 3*x* + 4*y* = 7

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

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જવાબ : 1:1

જવાબ : 1 : 2

If *α * be the length of the perpendicular from the origin on the line *x*/*a* + *y*/*b* = 1, then 1*/α*^{2 }= __________

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જવાબ : 1/a^{2} + 1/b^{2}

The point of intersection where the 5*x* + 3*y* − 20 = 0 perpendicular to the line 3*x* − 5*y* + 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 3^{rd} 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 5*x* + 3*y* − 7 = 0 and 2nd is __________

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જવાબ : 15*x* + 9*y* + 14 = 0

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

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જવાબ : *x* + 2*y* + 3 = 0

The value of λ for which the lines 3*x* + 4*y* = 5, 5*x* + 4*y* = 4 and *x* + λ*y* = 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 3*x* + 2*y* + 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 4*x* + 3*y* − 7 = 0 and 8*x* + 5*y* − 1 = 0 is __________

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જવાબ : 3*x* + 2*y* − 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*x *+ 3y + 10 = 0, 5*x* − 12*y* + 26 = 0 and 7*x*+ 24*y* − 50 = 0 is __________

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જવાબ : (0, 0)

3*x* + 4*y *+ 2 = 0 divides the distance between the line 3*x* + 4*y* + 5 = 0 and 3*x* + 4*y* − 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 5*x* + *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)

The equation of the altitude that is perpendicular to

Line

∴2-3+

Thus, the equation of the altitude from the vertex

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*.

∴ tan*θ*=1

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

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જવાબ : Let the given points be *A *(2*x*, −2), *B* (4, 2) and *C* (8, 10).

Slope of *AB* = 2+2/4-2*x *= 2/2-*x*

Slope of *BC* = 10-2/8-4 = 4/2 = 2

It is given that the points (2*x*, −2), (4, 2) and (8, 10) are collinear.

∴ Slope of *AB* = Slope of *BC*

⇒2/2-*x*=2

Hence, the value of

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* = *m*_{1} = 8-7/2+2 =1/4

Slope of *PQ* = *m*_{2 }= 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.

∴ *m*_{1}*m*_{2}=-1

Hence, the value of

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*.

∴ tan*θ*=1

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)

Solving

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 4*x* − 6*y* + 2 = 0, 2*x* + 2*y* = 6, 4*x* − 6*y* = 4 and 2*x* + 2*y* = 8 form a parallelogram.

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જવાબ : The given lines can be written as

*y*=2/3*x*+1/3 ... (1)

*y=-x+3* ... (2)

*y*=2/3*x*-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 : 2*x*+*y*+26=0

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

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

It passes through (2, 4).

∴6-16+*λ*=0⇒*λ*=10

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

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

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

It passes through (2, 2).

∴6-8+*λ*=0

Hence, the required line is 3

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*

This is the slope intercept form of the given line.

∴ Slope = -

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.

Math

- Math Book for CBSE Class 11
- Chemistry I Book for CBSE Class 11
- Chemistry II Book for CBSE Class 11
- Physics I Book for CBSE Class 11
- Physics II Book for CBSE Class 11
- Biology Book for CBSE Class 11

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.