*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 _______

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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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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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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-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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Answer :

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__________

Answer :

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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Answer :

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 __________

Answer :

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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Answer :

*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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Answer :

*a*^{2}

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*.

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 *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).

Answer :

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

⇒*θ*=tan^{-1}(1) = π/4

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

Answer :

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

⇒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*.

Answer :

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

⇒ 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).

Answer :

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

⇒*θ *= tan^{-1}(1) = 45^{0}

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

Hide | ShowAnswer :

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.

Answer :

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)

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

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* = *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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Answer :

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.

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.

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