Let A = {9,10,11,12,13} and let f : A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

 11th cbse ncert chapter 2 relations and functions miscellaneous exercise

12. Let A = {9,10,11,12,13} and let f : A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

9 = 3*3 therefore f(9)=3  

10= 2*5 therefore f(10)=5

11 =11 therefore f(11)=11

12 = 2 * 2 * 3 therefore f(12)=3

13 =13 therefore f(13)=13  

range of f = {3,5,11,13}

11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a
function from Z to Z? Justify your answer.

choose a =(-1), b=(-1)

a*b = (-1)*(-1)  = 1

a+b = (-1)+(-1) = (-2)

therefore ( 1, -2 ) belongs to f

now choose a =1 , b =1

a*b = 1*1 = 1

a+b = 1+1 =2

therefore (1,2) belongs to f

therefore the element 1 has more than one image under f namely 

(-2) and 2.

Therefore f is not a function .

 

10. Let A ={1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5), (2,9), (3,1), (4,5), (2,11)}
Are the following true?
(i) f is a relation from A to B
(ii) f is a function from A to B.
Justify your answer in each case. 

clearly  each element of f belongs to A x B. So f is a relation on A to B

both (2,9) and (2,11) belong to f 

so f has more than one image under f

so f is not a function.

 

=================================================

11th cbse ncert chapter 2 relations and functions miscellaneous exercise

12. Let A = {9,10,11,12,13} and let f : A → N be defined by f (n) = the highest prime factor of n. Find the range of f.

solution

 

11. Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a
function from Z to Z? Justify your answer.

solution

 

ncert cbse 11th mathematics chapter 11 conic section 

miscellaneous

 If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus

solution

2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high
and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?

solution

 

3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola.The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m. Find the length of a supporting wire attached to the roadway 18 m from the
middle.

solution  

4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre.Find the height of the arch at a point 1.5 m from one end. 

solution  

 5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.

solution

Find the area of the triangle formed by the lines joining the vertex of the parabola
x^2 = 12y to the ends of its latus rectum.  

solution 

7. A man running a racecourse notes that the sum of the distances from the two flagposts from him is always 10 m and the distance between the flag posts is 8 m.
Find the equation of the posts traced by the man

solution 

8. An equilateral triangle is inscribed in the parabola (y^2) = 4 ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

 solution 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

ncert cbse chapter 10  straight lines exercise 10.2 

 17.The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear
relationship between selling price and demand, how many litres could he sell
weekly at Rs 17/litre?

choose x as selling price per litre in rupees and y as demand in litres 

the line passes through (14,980 ) and (16,1220)

using 2 point form equation of the line is

[(y-980)/(1220-980)] = [(x-14)/(16-14)]

(y-980)/240 = (x-14)/2

y-980 = 120(x-14)

y-980 = 120x -1680 

 

y = 120x-1680+980

y = 120x -700

 

now put x = 17

y = 120*17 -700 

y = 1340 litres

 

13. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9. 

let the intercepts be a and b

given a+b = 9

b = (9-a)

using intercept form of straight line

required equation is

[x/a] + [y/b] =1

change  b = (9-a) to get

[x/a] + [y/ (9-a)] =1--------------(1)

this passes through (2,2) 

so 

[2/a] + [2/ (9-a)] =1

 

[ 2(9-a)  +2a ] / [a(9-a)] = 1

[18 ]/ [a(9-a)] = 1

18 = [a(9-a)]

18 = 9a -(a^2) 

(a^2)  - 9a +18 = 0

(a-3)(a-6) =0

a = 3  or a=6

use in eqn(1) to get

2x+y =6  or x+2y =6

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

 

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

solution

 

10. The line through the points (h, 3) and (4, 1) intersects the line 

7 x − 9 y − 19 = 0 at right angle. Find the value of h. 

solution

 

 

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

solution  

 

exercise 10.2

19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find
equation of the line.

solution

17.The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear
relationship between selling price and demand, how many litres could he sell
weekly at Rs 17/litre?

solution

13. Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.  

solution 

 

12.Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

 solution

 

11.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.   

 solution 

10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6). 

solution

 

9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

solution

 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

 ncert cbse chapter 10  straight lines exercise 10.2

19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find
equation of the line.

let the intercepts of the line be a and b so that the line cuts the x axis and y axis at

(a,0)  and (0,b) respectively.

using section formula with ratio 1:2 ,

 ( ( 0 + 2a )  / (1+2) ,  (1b+0) / (2+1) ) = (h,k) by given condition

 so

2a/3 = h  and b/3 = k

a =3h/2  ; b= 3k

use equation of intercept form

(x/a) + (y/b) = 1

[x / (3h/2)] + [y / (3k)] = 1

2kx + hy = 3hk

 

10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6). 

slope of the line passing through (2, 5) and (–3, 6)

is m1 = [6-5]  / [(-3) -2] = 1 /(-5)

 

condition for perpendicular lines is

m1 * m2 = (-1)

[1 /(-5)] * m2 = (-1)

so 

m2 = 5

is the slope of the required line.

required line passes through (-3,5) with slope 5

use point slope form to get the

equation of line as

y-5 =5[x+3] 

y-5 = 5x+15

5x-y+20 = 0

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

 

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

solution

 

10. The line through the points (h, 3) and (4, 1) intersects the line 

7 x − 9 y − 19 = 0 at right angle. Find the value of h. 

solution

 

 

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

solution  

 

exercise 10.2

19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find
equation of the line.

solution

12.Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

 solution

 

11.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.   

 solution 

10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6). 

solution

 

9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

solution

 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means 

 

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

 ncert cbse chapter 10  straight lines exercise 10.2

 12. Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

 

Let the equation of the line be {x/a}  +{y/b} = 1

given intercepts are equal

put b = a

equation changes to {x/a}  +{y/a} = 1

so 

x + y = a --------------(1) 

(1) passes through (2,3)

so 2 + 3 = a

or a = 5

substitute back in (1)

equation changes to x + y = 5

 

11.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line. 

slope of the line joining (1, 0) and (2, 3) is

m1 = [3-0]/[2-1] = 3

using the condition for perpendicular lines

m1 * m2 = (-1)

3 * m2 = (-1)

m2 = (-1) / 3 is the slope of the required line

using section formula  the point which divides the line joining  (1, 0) and (2, 3)  in the ratio 1: n is given by

(  [(n+2) / (n+1)] , [(0 + 3) / (n+1)] ) = (  [(n+2) / (n+1)] , [(3) / (n+1)] )

required line passes through (  [(n+2) / (n+1)] , [(3) / (n+1)] ) 

with slope m2 = (-1) / 3 

 

using point slope form

equation  is 

[y - [(3) / (n+1)] ] = [(-1) / 3 ] *[x - [(n+2) / (n+1)] ]

3[n+1]y - 9 = -(n+1)x +(n+2)

or 

(n+1)x + 3[n+1]y = n+11

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

 

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

solution

 

10. The line through the points (h, 3) and (4, 1) intersects the line 

7 x − 9 y − 19 = 0 at right angle. Find the value of h. 

solution

 

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

solution  

 

exercise 10.2

12.Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

 solution

 

11.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. Find the equation of the line.   

 solution 

 

9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

solution 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

 ncert cbse chapter 10  straight lines exercise 10.2

9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

using midpoint formula

 P (2, 1), Q (–2, 3)

mid point of PQ is M =[(2+(-2))/2  , (1+3)/2]  = (0,2)

median through R(4,5) passes through M(0,2)

using two point form equation of the median is

(y-5) / (2-5) = (x-4) / (0-4)

(y-5) / (-3) = (x-4) / (-4)

(-4) (y-5) = (-3)(x-4)

-4y+20 = -3x+12

3x-4y+8 = 0

exercise 10.3 straight lines

10. The line through the points (h, 3) and (4, 1) intersects the line 

7 x − 9 y − 19 = 0 at right angle. Find the value of h. 

slope of the line joining (h, 3) and (4, 1) is

m1 = (1-3) / (4-h) = (-2) / (4-h)

slope of the line 7 x − 9 y − 19 = 0

is m2 = -{coefficient of x]  / [coefficient of  y] = -(7) / (-9) = [7/9]

condition for perpendicular lines is 

m1 * m2 = (-1)

[ (-2) / (4-h) ] * [7/9 ] = (-1)

14 = 36 -9h

9h = 22

h = 22/9

 

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

 

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

solution

 

10. The line through the points (h, 3) and (4, 1) intersects the line 

7 x − 9 y − 19 = 0 at right angle. Find the value of h. 

solution

 

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

solution  

 

exercise 10.2

9. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the
median through the vertex R.

solution 

 

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13. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

 ncert cbse chapter 10 exercise 10.3

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

Using midpoint formula

midpoint of the line segment joining  (3, 4) and (–1, 2) is

[(3+(-1))/ 2  , (4+2)/2] = (1,3)

 

slope of the given line segment is   [2-4]/ [ (-1) -3] = (-2)/(-4) =1/2

 using condition for perpendicular lines m1*m2 = (-1)

so slope of the right bisector is (-2)

right bisector passes throught (1,3) with slope [-2]

using point slope form

y-y1 = m [x-x1]

equation of right bisector is

y-3 = [-2] [x-1]

y-3  = -2x +2

2x+y-5 =0

or 2x +y = 5

 

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

 

using slope of ax+by+x = 0 is given by m = (-a) / b

slope of x – 7y + 5 = 0 is  m1 = (-1) / (-7) = 1/7

using condition of perpendicular lines m1 * m2 = -1

slope of the required line is m2 = (-7)

x intercept 3 means that the line passes through (3,0)

so required line passes through (3,0) with slope (-7)

using point slope form

y - y1 = m [x-x1]

y -0 = (-7)[x -3]

y = -7x +21

7x+y = 21

 

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 13. Find the equation of the right bisector of the line segment joining the points

 (3, 4) and (–1, 2).

solution

8. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having
x intercept 3.

solution 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other mean

17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

 17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

 

B (4, –1) and C (1, 2)

using two point form equation of BC is 

{(y-(-1)) / [2-(-1)]} = {(x-4) / (1-4)} 

[y+1] / 3 = [x-4] / (-3)

y+1 = -x+4

x+y-3 = 0--------------(1)

 

line perpendicular to ax+by+c=0 is of the form bx-ay+k=0

 

altitude through A is perpendicular to BC

so the form of the altitude is

x-y+k=0--------------------(2)

This passes through A(2,3)

so 

2-3+k =0

k = 1

so required altitude through A is

x - y +1 =0 using k = 1 in (2)

solving  equation of altitude with that of BC (1)

 

x -  y = (-1)

x + y = 3

solving 

2x = 2

x = 1

resubstitute

y =2

point of intersection is P(1,2)

A is (2,3)

 

using distance formula

length of the altitude is

sqrt[ {(2-1)^2} + {(3-2)^2}] 

=sqrt[2]

we can also use perpendicular distance formula to find the distance between

 A(2,3)  and the line BC { x + y = 3 } for finding the length of the altitude.

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

Given line is 3x – 4y – 16 = 0

line perpendicular to ax+by+c=0 is of the form bx-ay+k=0

line perpendicular to given line is

-4x-3y+k = 0

This passes through  (–1, 3) if

-4(-1) - 3(3) + k =0

k =5

so perpendicular line is

-4x-3y+5 = 0

or 4x + 3y  = 5

given equation is

3x - 4y = 16

solving

[4x + 3y  = 5  ] *3

[3x - 4y = 16  ]  *4

 

12x+9y =15

12x-16y=64

 

subtracting

25y =(-49) 

y = [(-49) / 25]

resubstitute and solve for x

3x -4 [(-49) / 25] = 16

3x =16 -[196/25]

3x =204/25

x =68/25

 

foot of the perpendicular is [(68/25) , (-49/25)]

 

 

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line [sqrt(3)] x + y + 2 = 0.

solution 

 

 Find the values of k for which the line 

(k–3) x – (4 –( k^ 2) ) y + (k^2) –7k + 6 = 0 is

(a) Parallel to the x-axis,
(b) Parallel to the y-axis,
(c) Passing through the origin.

 solution  

ncert cbse chapter 10 exercise 10.3

  17. In the triangle ABC with vertices A (2, 3), B (4, –1) and C (1, 2), find the equation and length of altitude from the vertex A.

solution

 

14. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the
line 3x – 4y – 16 = 0.

solution

 

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means

 

Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and – 6, respectively.

 ncert cbse chapter 10 straight lines miscellaneous exercise

3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

let the intercepts be a and b 

 

given a+b =1 therefore b = (1-a)

given ab = (-6)  use b = (1-a)

a(1-a) = (-6)

a - (a^2) = (-6)

(a^2) - a -6 =0

(a-3)(a+2) = 0

a =3 ; a =(-2)

If a =3, b = 1-a = 1-3 = (-2)

If a =(-2) , b = 1 -a = 1 - (-2) =3

If a = 3 , b =(-2), equation of line is [x/2] + [y/b] =1

[x/3] + [y/(-2)] = 1

2x -3y =6

 

If a = (-2) , b =3 , equation of line is [x/2] + [y/b] =1

[x/(-2)] + [y/3] = 1

-3x+2y =6 or 3x-2y+6=0 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

let the point on the y axis be (0,k)

Given equation is [x/3] + [y/4]=1

re arrange

  4x+3y-12 =0

use formula for perpendicular distance  

for (0,k) to the line 4x+3y-12 =0

distance =|4(0)+3(k)-12|  / sqrt[(4^2)+(3^2)]

= |3(k)-12|  / 5

use given distance

4 =  |3(k)-12|  / 5

 |3(k)-12|  = 20

changing absolute value

3(k)-12 = 20  or 3(k)-12 = (-20)

we get

3k = 32  or 3k = -8

k = 32/3  or k =(-8/3)

required point is

(0 , 32/3) or (0, (-8/3) )

 

 

=================================================

ncert cbse chapter 10 straight lines miscellaneous exercise

24.  A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

solution

 

22. A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

 

solution 

 21. Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 =0

solution

 

19. If the lines y = 3x +1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

solution

18.Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

solution 

 

17. The hypotenuse of a right angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle 

 solution

 14. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the 

line x + y = 4 ?

solution

12.Find the equation of the line passing through the point of intersection of the lines 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes

solution

 

11. Find the equation of the lines through the point (3, 2) which make an angle of 45 degrees with the line x – 2y = 3.

solution

 

8. Find the area of the triangle formed by the lines y – x = 0, x + y = 0 

and x – k = 0

solution

9. Find the value of p so that the three lines 3x + y – 2 = 0, px + 2 y – 3 = 0 and
2x – y – 3 = 0 may intersect at one point.

 solution

6. Find the equation of the line parallel to y-axis and drawn through the point of
intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

solution

 

4. What are the points on the y-axis whose distance from the line
[x/3] + [y/4]=1 is 4 units.

solution

 

 3. Find the equations of the lines, which cut-off intercepts on the axes whose sum
and product are 1 and – 6, respectively.

solution

 

ncert cbse chapter 9 sequences and series miscellaneous exercise

 

 32.

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on the second day. 4 more workers dropped out on the third day and so on. It took 8 more days to finish the work. Find the number of days in which the work was finished

solution

 

disclaimer:
There is no guarantee about the data/information on this site. You use the data/information at your own risk. You use the advertisements displayed on this page at your own risk.We are not responsible for the content of external internet sites. Some of the links may not work. Your internet usage may be tracked by the advertising networks and other organizations using tracking cookie and / or using other means