NCERT Solutions for Class 11 Maths Miscellaneous Exercise Straight Lines - Future Study Point

# NCERT Solutions for Class 11 Maths Miscellaneous Exercise Straight Lines

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Straight Lines are created by an expert of Future Study Point for helping the online students for doing their homework and preparation of class 11 maths exam. NCERT Solutions for Class 11 Maths Miscellaneous Exercise Straight Lines are the extract of all the exercises from 10.1 to 10.3 of the chapter 10,so it is most important exercise for the preparation of class 11 maths exam of CBSE board.

Exercise 10.1-Straight Lines

Exercise 10.3 -Straight Lines

Q1.Find the values of k for which the line (k -3)xย  -(4 -kยฒ)y +kยฒ-7k +6 = 0 is

(a) Parallel to the x axis

(b)Parallel to the y axis

(c)Passing through the origin

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Ans. The given line is (k -3)xย  -(4 -kยฒ)y +kยฒ-7k +6 = 0

-(4 -kยฒ)y =-(k -3)xย  -kยฒ +7k -6

-(4 -kยฒ)y =-(k -3)xย  -kยฒ +7k -6

Comparing the equation with y = mx +c

Slope (m) of the given equation is (k-3)/(4 -kยฒ)

The equation of the x axisย  is y= 0,slope of x axis is 0

We know the slope of two parallel lines are equal to each other

Therefore

(k-3)/(4 -kยฒ) = 0

k -3 = 0

k = 3

Hence if the given line is parallel to x axis then the value of is k =3

Q2.ย Find the values of ฮธ and p, if the equation x cos ฮธ + y sin ฮธ = p is the normal form of the line โ3x + y + 2 = 0

Ans. The given equation is โ3x + y + 2 = 0

Comparing it to normal form of the equation x cos ฮธ + y sin ฮธ = p

โ3x + y=- 2

-โ3 x -y = 2

Dividing both sides by โ[(-โ3)ยฒ +(-1)ยฒ]= โ(3 +1)=2

Since cos and sin are negative,therefore ฮธ lies in third quadrant

cosฮธ =cos(ฯ +ฯ/6), sinฮธ =sin(ฯ +ฯ/6)

cosฮธ =cos7ฯ/6,sinฮธ =sin7ฯ/6

ฮธ =7ฯ/6 and p =1

Solutions of Q1 and Q2

Q3.Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and โ6, respectively.

Ans. Let x intercept of the equation is x and y intercept of the equation is b

The sum of both intercept is given 1 and product is -6

a + b = 1….(i)

ab = -6……(ii)

Putting the value of b = (1-a) from equation (i) in equation ( ii)

a(1-a) = -6

a -aยฒ +6 =0

aยฒ -a -6 =0

aยฒ -3a +2a -6=0

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

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

a=3,-2

For a = 3, b=-2 and for a =-2, b=3

If the x itercept,aย  and y intercept ,bย  then equation is written as follows

If a =3 and b= -2 then the equation is as follows

If a =-2 and b= 3 then the equation is as follows

Therefore the required equations are 2x -3y -6=0 and -3x +2y -6 =0

Q4.What are the points on theย y-axis whose distance from the lineย x/3 + y/4 = 1 is 4 units?

Ans. Let the point on y axis is (0,a) whose distance from the line x/3 + y/4 = 1 is 4 units

The given equation is x/3 + y/4 = 1

Writting the given equation into the general form

4x +3y -12 = 0

The distance (d) of a lineย  Ax +By + C = 0 from a point (x1,y1)

The distance of the point (0,a) from the line 4x +3y -12 = 0 ,where A =4, B =3 and C = -12

Since d is given to us = 4 units

20ย  = ยฑ(3a -12)

In the case 20 = 3a -12โ 3a =32 โ a =32/3

In the case 20 =-(3a -12) โ20 =-3a +12โa =-8/3

Therefore the required points on y axis are (0,32/3) and (0,-8/3)

Q5.Find the perpendicular distance from the origin to the line joining the pointsย

(cos ฮธ,sinฮธ) and (cos ฮฆ,sinฮฆ)

Ans. The equation of a straight line passing through a point (x1,y1) is given as follows

(y -y1) = m(x -x1),where m is the slope

The slope of the line(m) joining two points (cos ฮธ,sinฮธ) and (cos ฮฆ,sinฮฆ) is as follows

Also the line is passing through (cos ฮธ,sinฮธ),so x1=cos ฮธ and y1=sin ฮธ

(y -sinฮธ)(cosฮฆ -cosฮธ)ย  = (sinฮฆ -sinฮธ)(x -cosฮธ)

y(cosฮฆ -cosฮธ) -sinฮธ(cosฮฆ -cosฮธ) =sinฮฆ(x -cosฮธ)-sinฮธ(x -cosฮธ)

y(cosฮฆ -cosฮธ) -sinฮธcosฮฆ +sinฮธcosฮธ = xsinฮฆ -sinฮฆ cosฮธ -xsinฮธ +sinฮธcosฮธ

y(cosฮฆ -cosฮธ) +x(sinฮธ -sinฮฆ)ย  +(sinฮฆ cosฮธ-sinฮธcosฮฆ) =0

y(cosฮฆ -cosฮธ) +x(sinฮธ -sinฮฆ)ย  +sin(ฮฆ-ฮธ) = 0

Rearranging into the general form

x(sinฮธ -sinฮฆ)ย  +y(cosฮฆ -cosฮธ) + sin(ฮฆ-ฮธ) =0

The distance(d) of the point (x1,y1)ย  from the line Ax +By +C =0

Therefore the distance (d) from the origin(0,0) to the line x(sinฮธ -sinฮฆ)ย  +y(cosฮฆ -cosฮธ) -sin(ฮธ -ฮฆ) =0

Q6.Find the equation of the line parallel toย y-axis and drawn through the point of intersection of the lines xย โ 7yย + 5 = 0 and 3xย +ย yย = 0.

Ans. A line which is parallel to Y axis creats 90ยฐ angle with positive side of X axis

Therefore the slope ,m of the line is = tan 90ยฐ = sin 90ยฐ/cos 90ยฐ =1/0

Also the line is passing through the point of intersection of both lines xย โ 7y + 5 = 0…(i) and 3xย +ย y = 0…..(ii)

For solving both of the lines ,multiplying equation (i) by 3,we get equation (iii)

3x – 21y + 15 =0…..(iii)

Subtracting equation (iii) from equation (ii)

22y -15 =0

22y = 15

y = 15/22

Putting the value of y in equation (i)

3x + 15/22 = 0

3x = -15/22

x = -15/66 = -5/22

Therefore both of the given lines intersects each other at (15/22,-5/22)

The equation of the line which is passing through a point (x1,y1) with slope m is given by

y – y1=m (x -x1)

The equation of the line which is passing through (15/22,-5/22) with slope 1/0

y – 15/22 = (1/0)(x +5/22)

x + 5/22 =0

x = -5/22

Hence the required equation of the line is -5/22

Q7.Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis.

Ans. Let the given line meets the Y axis at (0,y),wherefrom we have to find out the equation of perpendicular line

Writting the given line x/4 + y/6 = 1 into the x intercept form y = mx +c

3x + 2y = 12

2y = -3x +12

y = (-3/2)x + 6

On comparing the standard form

m = -3/2

We know

The product of slopes of two perpendicular line =-1

Slope of perpendicular line to given line = -1/Slope of the given line= -1/(-3/2) = 2/3

Since the given lineย  is passing through (0,y) ,so putting x =0 in the given equation

y/6 =1

y = 6

The equation of perpendicular line drawn from (0,6) is given as

y -6 = (2/3)( x -0)

3y -18 =2x

2x -3y +18 =0

Hence required equation of the line is 2x -3y +18 =0

Q8. Find the area of the triangle formed by the lines yย โย xย = 0,ย xย +ย yย = 0 andย xย โย kย = 0.

Ans. Let’s find the verices of the triangle by solving these equations

The given equations are yย โย xย = 0….(i) xย +ย y = 0….(ii) and xย โย k = 0…..(iii)

Solving equations (i) and (ii),we get y = 0 and x =0,so one of the vertex is (0,0)

Solving equations (ii) and (iii),we get y = -k and x =k,so one of the vertex is (k,-k)

Solving equations (i) and (iii),we get y = k and x =k,so one of the vertex is (k,k)

The area of the triangle having the verticesย (x1,y1)(x2,y2) andย (x3,y3) is given as

Area of ฮ = (1/2)[ (x1(y2– y3) + x2(y3– y1) +x3(y1– y2) ]

Area of the triangle whose vertices are (0,0),(k,-k) and (k,k)

Area of ฮ = (1/2) [0(-k-k) +k(k-0) + k(0 +k)]

=(1/2) [0 + kยฒ + kยฒ]

(1/2) .2kยฒ = kยฒ

Hence area of the required triangle is kยฒ units.

Q9.Find the value ofย pย so that the three lines 3xย +ย y – 2 = 0, pxย + 2y -3 = 0 and 2x –ย y -3 = 0 may intersect at one point.

Ans.The given lines are 3xย +ย y – 2 = 0…(i) pxย + 2y -3 = 0 …..(ii) and 2x –ย y -3 = 0 ….(iii)

Solving the equations (i) and (iii),we get the point of intersection of (i) and (iii) equations

Adding equation (iii) to equation (i)

5x -5 =0

5x =5

x =1

Putting the value of x in equation (i)

3ร1 + y -2 =0

3 + y -2 =0

y +1 =0

y =-1

The point of intersection of (i) and (iii) line is (1,-1),since it is given that three lines intersect on the same point,therefore the point (1,-1) should satisfy the (ii) equation

Putting x =1 and y =-1

pร1+2ร-1-3=0

p -2 -3 =0

p-5 =0

p =5

Hence the required value of p is 5

Q10. If three lines whose equations are y = m1x + c1, y = m2x + c2ย and y = m3x + c3ย are concurrent, then show that m1ย (c2 – c3) + m2ย (c3 – c1) + m3ย (c1 – c2) = 0.

Ans. The given equations are y = m1x + c1…..(i) y = m2x + c2ย ……(ii) and y = m3x + c3 …….(iii)

Solving equations (i) and equation (ii),for this,substracting equation (ii) from equation (i)

0 = x(m1– m2) +(c1ย  -c2)

x = (c2ย  -c1)/(m1– m2)

Putting the value of x in equation (i)

y = m1x + c1

y ย = m1ย (c2ย  -c1)/(m1– m2)+ c1

Since lines (i),(ii) and (iii) are concurrent therefore the line (i) and (ii) intersect each other at the pointsย  [(c2ย  -c1)/(m1– m2), m1ย (c2ย  -c1)/(m1– m2)+ c1] should satisfy the equation (iii)

m1ย (c2ย  -c1)/(m1– m2)+ c1=m3 (c2ย  -c1)/(m1– m2)+c3

m1ย (c2ย  -c1)/(m1– m2)=m3 (c2ย  -c1)/(m1– m2)+c3-c1

m1ย (c2ย  -c1)=m3 (c2ย  -c1)+(c3-c1)(m1-m2) =0

m1c2– m1c1ย = m3c2–ย m3c1 + m1c3– m2c3– m1c1 + m2c1

m1c2ย = m3c2–ย m3c1 + m1c3– m2c3+ m2c1

m1c2 ย -m3c2+m3c1 – m1c3+m2c3 – m2c1=0

m1ย (c2ย โ c3) + m2ย (c3ย โ c1) + m3ย (c1ย โ c2) = 0,Hence proved

Q11.Find the equation of the lines through the point (3, 2) which make an angle of 45ยฐ with the lineย x – 2yย = 3.

Ans.Let the slope of the line which is passing through the given point (3,2) =m1

Rewritting the given equation of the line into the slope intercept form y =mx +c

-2y = -x + 3

y = -x/-2 + 3/-2

y =x/2ย  -3/2

On comparing with y = mx +c

Slope of the line is = 1/2

Let it isย m2=1/2

The angle between both lines given to us = 45ยฐ

The relationship between slopes andย  angle between both lines is given as

Putting the values ofย m1 ,m2ย and tan ฮธ

It is written as

tan 45ยฐ = ยฑ(1- 2m1)/(2 +m1)

Taking positive root

1 = (1- 2m1)/(2 +m1)

2 +m1= 1- 2m1

3m1ย = 1-2 =-1

m1= -1/3

Taking negative root

1=-(1- 2m1)/(2 +m1)

2 +m1ย = -1+2m1

m1= 3

Therefore equations of the line passing through the point (3,2) are

y -2 = (-1/3)(x -3) and y -2 = 3(x -3)

3y -6 = -x +3 and y-2 = 3x -9

x +3y -9 =0 and 3x -y -7 =0

Hence required equations of the line are x +3y -9 =0 and 3x -y -7 =0

Q12. Find the equation of the line passing through the point of intersection of the lines 4xย + 7y – 3 = 0 and 2x – 3yย + 1 = 0 that has equal intercepts on the axes.

Ans. The given equations given to us are 4xย + 7y – 3 = 0…(i) and 2x – 3y + 1 = 0….(ii)

Multiplying the equation (ii) by 2,we get equation (iii)

4x -6y +2 =0…(iii)

Substracting equation (iii) from equation (i)

13y -5 = 0

y = 5/13

Putting this value of y in equation (i)

4x + 7 ร5/13 -3 =0

4x = 3 -35/13 = (39 -35)/13 =4/13

x = 1/13

So, point of intersection of both given line is (1/13,5/13)

The required line is passing through (1/13,5/13) with equal interceps of both axis.

Let the both intercepts of the axis are a

Therefore

x/aย  + y/a = 1

x +y = a

It should satisfy (1/13, 5/13)

1/13ย  +5/13 =a

a = 6/13

Hence the required equation is

x + y = 6/13

13x + 13y = 6

Q13.Show that the equation of the line passing through the origin and making an angle ฮธย with the line y = mx + c isย

Ans. We have to find out equation of the line which is passing through the origin,so supossing its slopeย m1ย the equation is givenย by

y -0 =ย m1(x -0)

m1ย = y/x

The given equation is y = mx + c,therefore its slope,=m2= m

The relationship between slopes andย  angle between both lines is given as

Putting the values ofย m1 ,m2ย and tan ฮธ

tan ฮธ = ยฑ(m- y/x)/( 1 + ym/x)

Taking the positive roots

tan ฮธ =(m- y/x)/( 1 + ym/x)

tan ฮธ + (y/x). m tan ฮธ = m – y/x

y/xย  + (y/x). m tanฮธย  = m – tan ฮธ

(y/x) (1 + m tanฮธ) = m – tan ฮธ

y/xย  = (m – tan ฮธ)/( 1 + m tan ฮธ)….(i)

Similarly we can get by taking tan ฮธ = -(m- y/x)/( 1 + ym/x)

y/x = (m + tan ฮธ)/( 1 – m tan ฮธ)….(ii)

From equation (i) and (ii) we get

Q14.In what ratio ,the line joining ,(-1,1) and (5,7) is divided by the line x + y =4.

Ans. Let, the line joining,(-1,1) and (5,7) is divided by the line x +y =4 in the ratio of m : n

So, let’s get point of intersection of the line joining (-1,1) and (5,7) and the line (x +y =4)

The slope of the line joining the given point is ,m =(7-1)/(5+1) =6/6 =1

The equation of the line passing through (-1,1) is given as

(y-1) = 1ร(x +1)

y-1 = x+1

y -x =2……(i)

The given equation is x +y =4….(ii)

Solving both equation (i) and (ii),we get y=3 and x=1

Therefore point of intersection of both lines (x +y=4) and (y-x =2) is (1,3)

The point (1,3) divides the line segment joining two points (-1,1) and (5,7) ย  ย in the ratio of m:n

Applying the section formula

1 = (5m -n)/(m+n)

m +n = 5m -n

4m =2n

m : n = 2 ;4

m : n =1 : 2

Or

3 =(7m +n)/(m+n)

3m +3n = 7m +n

-4m = -2n

m : n = 2 : 4 = 1 : 2

Therefore the given line divides the line segment joining the given points in the ratio of 1 :2

Q15.Find the distance of the line 4x +7y +5=0 from the point (1,2) along the line 2x -y =0.

Ans. The given lines are 4x +7y +5=0 ….(i) and 2x -y =0…..(ii)

Let’s find the distance of equation (i) from the point (1,2) which is on the line 2x -y =0.

The distance of the line (i) from (1,2) =Distance of the point (1,2) and the point of intersection line (i) and (ii)

Therefore solving both equation (i) and (ii)

Multiplying equation (ii) by 2 ,we get equation (iii)

4x -2y =0….(iii)

Substracting equation (ii) from equation (i)

9y +5 =0

y =-5/9

Putting the value of y in equation (ii)

2x -(5/9) =0

2x +5/9 =0

x =-5/18

Therefore point of intersection of the line (i) and (ii) is (-5/18,-5/9)

So, the distance(d) between the point (1,2) and (-5/18,-5/9)

d = โ[(1+5/18)ยฒ+ (2+5/9)ยฒ]

= โ[(23/18)ยฒ+ (23/9)ยฒ]

=โ[529/324+ 529/81]

=โ(529+ 2116)/324]=โ(2645/324) =23โ5/18

Hence the required distance of the line 4x +7y +5=0 from the point (1,2) is 23โ5/18

Q16.Find the direction in which a straight line must be drawn through the point (-1, 2) so that its point of intersection with the line xย +ย yย = 4 may be at a distance of 3 units from this point.

Ans.Let the slope of the straight line m,and its equation is y = mx +c

The line is passing through the point (-1,2)

Since c is a constant so,putting the values x=-1 and y =2

2 = m(-1) +c

c =2 +m

Putting the value of c =2 +m in the equation

y = mx + 2 +m

y = m(x +1) +2…..(i)

The given equation is

x +y =4 …….(ii)

Putting the value x = 4 -y from (ii)ย  ย in equation (i)

y = m(4-y +1) +2

y = 5mย  -my +2

y+ my = 5m +2

y(1 +m) = 5m +2

y = (5m +2)/(m+1)

Putting the value of y in equation (ii)

x + (5m +2)/(m+1) =4

x = 4-(5m +2)/(m+1) = (4m+4-5m-2)/(m+1) = (2-m)/(m+1)

So, the point of intersection of both lines is [(2-m)/(m+1),(5m +2)/(m+1)]

The distance between point of intersecton and the given point (-1,2) is given to us 3 units

3 =โ[ {(5m +2)/(m+1 ) -2}ยฒ +{(2-m)/(m+1) +1}ยฒ]

Squaring both sides

{(5m +2)/(m+1 ) -2}ยฒ +{(2-m)/(m+1) +1}ยฒ =9

(5m+2-2m-2)ยฒ/(m+1)ยฒ + (2-m+m+1)ยฒ/(m+1)ยฒ =9

(3m )ยฒ + 3ยฒ =9

9mยฒ +9 ย =9

9mยฒ =0

m =0

Hence slope of the line is 0 i.e the line is parallel to x axis

Q17.The hypotenuse of a right-angled triangle has its ends at the points (1, 3) and (-4, 1). Find the equation of the legs (perpendicular sides) of the triangle.

Ans. The given ends of the hypotenuse are (1,3) and (-4,1)

One of sideย  passes through (1,3) and the other passes through (-4,1)

The equation of the line which passes throughย (x1,y1) is given by

(y – y1) = m(x –x1)

Therefore the line that passes through (1,3) is

y -3 = m(x -1)……(i)

The line that passes through (-4,1) is

y -1 = -1/m(x +4)……(ii)(i.e the product of the slopes of two perpendicular line is -1)

The sides of the tringle can have any slope,so placing the specific value of m i.e m=0

Putting m =0 in equation (i) and in equation (ii)

y -3 = 0 โy= 3

y – 1= -1/0(x +4)

-x -4=0

x =-4

If equation of one side is y=3 then equation of other side is x =-4

Or if the slope of the line (i) is 1/m

y-3 = 1/m(x-1)

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

x-1=0โ x= 1

Then equation of second line is

y -1 =-m(x +4)

y-1 =0

y =1

If equation of one side is x =1 then equation of other side is y =1

Q18.Find the image of the point (3, 8) with respect to the line x + 3yย = 7 assuming the line to be a plane mirror.

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

The line x + 3y is the perpendicular bisector of the line segment PQ

The mid point of the line PQ = [(x+3)/2,(8+y)/2]

The mid point of PQ passes through the line x +3y =7,so it will satisfy its coordinates

(x+3)/2ย  + 3(8+y)/2 =7

x +3 +24 +3y =14

x + 3y + 13=0…….(i)

Since the line PQ โฅ (x +3y =7)

Therefore product of their slopes = -1

The slope of the given line (y =-x/3+7/3) is -1/3

The slope of PQ = (y-8)/(x -3)

โด (-1/3)(y-8)/(x-3) =-1

y -8 = 3x -9

3x -y -1 =0…..(ii)

Multiplying equation (i) by 3,we get equation (iii)

3x +9y +39 =0….(iii)

Substracting equation (iii) from equation (ii)

-10y -40 =0

y = -40/10 =-4

Putting the value y =-4 in equation (ii)

3x +4 -1 =0

3x = -3

x = -1

Hence the image of the given point (3,8) with respect to the given line is (-1,4)

Q19. If the lines y = 3x+ 1 and 2y= x + 3 are equally inclined to the line y = mx + 4, find the value of m.ย

Ans. The given equations are y = 3x+ 1….(i)ย  2y= x + 3 …..(ii) andย  y = mx + 4…..(iii)

Rearranging the equations into their slope-intercept form i.e y = mx +c

Slope of equation (i) = 3,Slope of equation (ii) =1/2 and Slope of equation (iii) =m

We are given that

The angle between y=3x +1 and the line y =mx +4 =The angle between 2y = x+3 and the line y = mx +4

(m-3)/(1 +3m) =ยฑ(2m-1)/(2 +m)

Taking the positive root

(m-3)/(1 +3m) =(2m-1)/(2 +m)

Cross multiplying it

(m-3)(2 +m) = (1 +3m)(2m -1)

2m +mยฒ-6 -3m = 2m -1 +6mยฒ-3m

5mยฒ +5 = 0

mยฒ +1 = 0

m= โ-1

m is not real ,so neglecting this value of m

(m-3)/(1 +3m) =-(2m-1)/(2 +m)

(m-3)(2 +m) = -(1 +3m)(2m -1)

2m +mยฒ-6 -3m = -2m +1 -6mยฒ+3m

7mยฒ-2m -7 =0

Hence the required value of m is (1ยฑ5โ2)/7

Q20.If sum of the perpendicular distances of a variable point P (x,ย y) from the lines x + y-5 = 0 and 3x – 2yย + 7 = 0 is always 10. Show that P must move on a line.

Let the perpendicular distances of the given lines from the point P(x,y) areย d1ย and

d1

The perpendicular distance(d) from a point P(x,y) to the given line Ax + By +C is given as

It is given to us that

d1+d1=10

โ13x +โ13 y -5โ13 + 3โ2 x -2โ2 y +7โ2 = 10โ26

x(โ13 +3โ2) + y(โ13 -2โ2) +7โ2-5โ13 –ย 10โ26 =0

The equation we have got is the equation of a line,so P(x,y) must move on it

Similarly we get the equation of another line by taking another sign

Q21.Find equation of the line which is equidistant from parallel lines 9xย + 6y – 7 = 0 and 3x + 2yย + 6 = 0.

Ans. Let the point (x,y) is an arbitrary point which is equidistant from parallel lines 9x +6y -7 =0 and 3x +2y +6 =0

The perpendicular distance(d) from a point (x,y) to the given line Ax + By +C is given as

According to the question

9x +6y -7 = ยฑ3(3x +2y +6)

Taking positive sign

9x +6y -7 = 9x +6y +18

-7 = 18,it is impossible

Taking negative sign

9x +6y -7 = -3(3x +2y +6)

9x +6y -7 = -9x -6y -18

18x +12y +11 =0

Hence the required equation of the line is 18x +12y + 11 =0

Q22. A ray of light passing through the point (1, 2) reflects on the x-axis at point B and the reflected ray passes through the point (5, 3). Find the coordinates of B.

Ans.

Let a ray AB incident on the x axis at the point (a,0) and reflected ray is BC

Drawing a normal BMโฅXX’

โ ABM =Incident angle=ฮฆ, โ CBM = Reflected angle=ฮฆ(i.e reflected angle =incident angle)

For finding the slope of the incident ray AB,let’s find the angle โ OBA

โ OBA = 180ยฐ-(ฮฆ+ฮฆ+ฮธ)= 180ยฐ-(2ฮฆ+ฮธ) = 180ยฐ-[2(90-ฮธ)+ฮธ)]= 180ยฐ-180 +2ฮธ-ฮธ =ฮธ

โ ABX’ = 180ยฐ-ฮธ

Therefore the slope of ray AB is = tan( 180ยฐ-ฮธ)=-tanฮธ

The slope of BC is = tanฮธ

Since AB is passes through (1,2) and (a,0) and BC passes through (a,0) and (5,3)

-tan ฮธ =(0-2)/(a-1)

tan ฮธ = 2/(a -1)……(i)

tan ฮธ =(3-0)/(5-a)= 3/(5-a)…..(ii)

2/(a-1) = 3/(5-a)

10 -2a =3a -3

5a =13

a = 13/5

Hence the required point on the X-axis is (13/5,0)

Q23. Prove that the product of the lengths of the perpendiculars drawn from the points [โ(aยฒ-bยฒ),0] and [-โ(aยฒ-bยฒ),0] to the line (x/a)cosฮธ +(y/b) sinฮธ =1 is bยฒ

Ans. We know the length of perpendicular (d) drawn from a point (x,y) to the line Ax +By +C =0 is given by

The given line is

(x/a)cosฮธ +(y/b) sinฮธ =1

Wriiting it in the standard form

(bcosฮธ)x + (asinฮธ)y -ab =0

Let the length of perpendiculars drawn from the given points to the given lines are

d1ย andย d2

Similarly

= bยฒ

Hence the required product of the perpendiculars is bยฒ

Q24. A person standing 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 equations is 6x – 7y + 8 = 0 in the lease time. Find the equation of the path that he should follow.

Ans. It is given that

2x – 3y + 4 = 0…. (i)

3x + 4y – 5 = 0…. (ii)

6x – 7y + 8 = 0…. (iii)

Here the person is standing at the junction of the paths represented by (i) and (ii)

By solving equations (i) and (ii) we get

x = – 1/17 and y = 22/17

Hence, the person is standing at point (- 1/17, 22/17)

We know that the person can reach path (iii) in the least time if he walks along the perpendicular line to (iii) from point (- 1/17, 22/17)

Here the slope of the line (iii) = 6/7

We get the slope of the line perpendicular to line (iii) = – 1/(6/7) = – 7/6

So the equation of line passing through (- 1/17, 22/17) and having a slope of – 7/6 is written as

By further calculation

6(17y – 22) = – 7(17x + 1)

By multiplication

102y – 132 = – 119x – 7

We get, 1119x + 102y = 125

Therefore, the path that the person should follow is 119x + 102y = 125

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## NCERT Solutions ofย  Science and Maths for Class 9,10,11 and 12

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

### NCERT Solutions of class 9 scienceย

 Chapter 1-Matter in our surroundings Chapter 9- Force and laws of motion Chapter 2-Is matter around us pure? Chapter 10- Gravitation Chapter3- Atoms and Molecules Chapter 11- Work and Energy Chapter 4-Structure of the Atom Chapter 12- Sound Chapter 5-Fundamental unit of life Chapter 13-Why do we fall ill ? Chapter 6- Tissues Chapter 14- Natural Resources Chapter 7- Diversity in living organism Chapter 15-Improvement in food resources Chapter 8- Motion Last years question papers & sample papers

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### NCERT Solutions of class 10 maths

 Chapter 1-Real number Chapter 9-Some application of Trigonometry Chapter 2-Polynomial Chapter 10-Circles Chapter 3-Linear equations Chapter 11- Construction Chapter 4- Quadratic equations Chapter 12-Area related to circle Chapter 5-Arithmetic Progression Chapter 13-Surface areas and Volume Chapter 6-Triangle Chapter 14-Statistics Chapter 7- Co-ordinate geometry Chapter 15-Probability Chapter 8-Trigonometry

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### NCERT solutions of class 10 science

 Chapter 1- Chemical reactions and equations Chapter 9- Heredity and Evolution Chapter 2- Acid, Base and Salt Chapter 10- Light reflection and refraction Chapter 3- Metals and Non-Metals Chapter 11- Human eye and colorful world Chapter 4- Carbon and its Compounds Chapter 12- Electricity Chapter 5-Periodic classification of elements Chapter 13-Magnetic effect of electric current Chapter 6- Life Process Chapter 14-Sources of Energy Chapter 7-Control and Coordination Chapter 15-Environment Chapter 8- How do organisms reproduce? Chapter 16-Management of Natural Resources

### Solutions of class 10 last years Science question papers

CBSE Class 10 – Question paper of science 2020 with solutions

CBSE class 10 -Latest sample paper of science

### NCERT solutions of class 11 maths

 Chapter 1-Sets Chapter 9-Sequences and Series Chapter 2- Relations and functions Chapter 10- Straight Lines Chapter 3- Trigonometry Chapter 11-Conic Sections Chapter 4-Principle of mathematical induction Chapter 12-Introduction to three Dimensional Geometry Chapter 5-Complex numbers Chapter 13- Limits and Derivatives Chapter 6- Linear Inequalities Chapter 14-Mathematical Reasoning Chapter 7- Permutations and Combinations Chapter 15- Statistics Chapter 8- Binomial Theorem ย Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### NCERT solutions of class 12 maths

 Chapter 1-Relations and Functions Chapter 9-Differential Equations Chapter 2-Inverse Trigonometric Functions Chapter 10-Vector Algebra Chapter 3-Matrices Chapter 11 – Three Dimensional Geometry Chapter 4-Determinants Chapter 12-Linear Programming Chapter 5- Continuity and Differentiability Chapter 13-Probability Chapter 6- Application of Derivation CBSE Class 12- Question paper of maths 2021 with solutions Chapter 7- Integrals Chapter 8-Application of Integrals

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