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

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Straight Lines

miscellaneous exercise chapter 10 class 11 maths

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Straight Lines

miscellaneous exercise chapter 10 class 11 maths

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)x  -(4 -k²)y +k²-7k +6 = 0 is

(a) Parallel to the x axis

(b)Parallel to the y axis

(c)Passing through the origin

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

Ans. The given line is (k -3)x  -(4 -k²)y +k²-7k +6 = 0

-(4 -k²)y =-(k -3)x  -k² +7k -6

-(4 -k²)y =-(k -3)x  -k² +7k -6

Comparing the equation with y = mx +c

Slope (m) of the given equation is (k-3)/(4 -k²)

The equation of the x axis  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 -k²) = 0

k -3 = 0

k = 3

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

Q2. Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0

Ans. The given equation is √3x + y + 2 = 0

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

√3x + y=- 2

-√3 x -y = 2

Dividing both sides by √[(-√3)² +(-1)²]= √(3 +1)=2

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

cosθ =cos(π +π/6), sinθ =sin(π +π/6)

cosθ =cos7π/6,sinθ =sin7π/6

θ =7π/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 –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 -a² +6 =0

a² -a -6 =0

a² -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,a  and y intercept ,b  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 the y-axis whose distance from the line 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 line  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

20  = ±(3a -12)

In the case 20 = 3a -12⇒ 3a =32 ⇒ a =32/3

In the case 20 =-(3a -12) ⇒20 =-3a +12⇒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 points 

(cos θ,sinθ) and (cos Φ,sinΦ)

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 θ,sinθ) and (cos Φ,sinΦ) is as follows

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

(y -sinθ)(cosΦ -cosθ)  = (sinΦ -sinθ)(x -cosθ)

y(cosΦ -cosθ) -sinθ(cosΦ -cosθ) =sinΦ(x -cosθ)-sinθ(x -cosθ)

y(cosΦ -cosθ) -sinθcosΦ +sinθcosθ = xsinΦ -sinΦ cosθ -xsinθ +sinθcosθ

y(cosΦ -cosθ) +x(sinθ -sinΦ)  +(sinΦ cosθ-sinθcosΦ) =0

y(cosΦ -cosθ) +x(sinθ -sinΦ)  +sin(Φ-θ) = 0

Rearranging into the general form

x(sinθ -sinΦ)  +y(cosΦ -cosθ) + sin(Φ-θ) =0

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

Therefore the distance (d) from the origin(0,0) to the line x(sinθ -sinΦ)  +y(cosΦ -cosθ) -sin(θ -Φ) =0

 

Q6.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.

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

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

Also the line is passing through the point of intersection of both lines x – 7y + 5 = 0…(i) and 3x + 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 line  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 y – x = 0, x + y = 0 and x – k = 0.

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

The given equations are y – x = 0….(i) x + y = 0….(ii) and x – 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 vertices (x1,y1)(x2,y2) and (x3,y3) is given as

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

=(1/2) [0 + k² + k²]

(1/2) .2k² = k²

Hence area of the required triangle is k² units.

Q9.Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y -3 = 0 and 2x – y -3 = 0 may intersect at one point.

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

3×1 + 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

p×1+2×-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 + c2 and y = m3x + c3 are concurrent, then show that m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0.

Ans. The given equations are y = m1x + c1…..(i) y = m2x + c2 ……(ii) and y = m3x + c3 …….(iii)

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

0 = x(m1– m2) +(c1  -c2)

x = (c2  -c1)/(m1– m2)

Putting the value of x in equation (i)

y = m1x + c1

y  = m1 (c2  -c1)/(m1– m2)+ c1

Since lines (i),(ii) and (iii) are concurrent therefore the line (i) and (ii) intersect each other at the points  [(c2  -c1)/(m1– m2), m1 (c2  -c1)/(m1– m2)+ c1] should satisfy the equation (iii)

m1 (c2  -c1)/(m1– m2)+ c1=m3 (c2  -c1)/(m1– m2)+c3

m1 (c2  -c1)/(m1– m2)=m3 (c2  -c1)/(m1– m2)+c3-c1

m1 (c2  -c1)=m3 (c2  -c1)+(c3-c1)(m1-m2) =0

m1c2– m1c= m3c2– m3c1 + m1c3– m2c3– m1c1 + m2c1

m1c2 = m3c2– m3c1 + m1c3– m2c3+ m2c1

m1c2  -m3c2+m3c1 – m1c3+m2c3 – m2c1=0

m1 (c2 – c3) + m2 (c3 – c1) + m3 (c1 – c2) = 0,Hence proved

Q11.Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 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/2  -3/2

On comparing with y = mx +c

Slope of the line is = 1/2

Let it is m2=1/2

The angle between both lines given to us = 45°

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

Putting the values of m1 ,mand tan θ

It is written as

tan 45° = ±(1- 2m1)/(2 +m1)

Taking positive root

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

2 +m1= 1- 2m1

3m1 = 1-2 =-1

m1= -1/3

Taking negative root

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

2 +m1 = -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 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

Ans. The given equations given to us are 4x + 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 ×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/a  + y/a = 1

x +y = a

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

1/13  +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 θ with the line y = mx + c is 

Ans. We have to find out equation of the line which is passing through the origin,so supossing its slope m1 the equation is given by

y -0 = m1(x -0)

m1 = y/x

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

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

Putting the values of m1 ,mand tan θ

tan θ = ±(m- y/x)/( 1 + ym/x)

Taking the positive roots

tan θ =(m- y/x)/( 1 + ym/x)

tan θ + (y/x). m tan θ = m – y/x

y/x  + (y/x). m tanθ  = m – tan θ

(y/x) (1 + m tanθ) = m – tan θ

y/x  = (m – tan θ)/( 1 + m tan θ)….(i)

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

y/x = (m + tan θ)/( 1 – m tan θ)….(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) = 1×(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)    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 = √[(1+5/18)²+ (2+5/9)²]

= √[(23/18)²+ (23/9)²]

=√[529/324+ 529/81]

=√(529+ 2116)/324]=√(2645/324) =23√5/18

Hence the required distance of the line 4x +7y +5=0 from the point (1,2) is 23√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 x + y = 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)   in equation (i)

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

y = 5m  -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 =√[ {(5m +2)/(m+1 ) -2}² +{(2-m)/(m+1) +1}²]

Squaring both sides

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

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

(3m )² + 3² =9

9m² +9  =9

9m² =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 side  passes through (1,3) and the other passes through (-4,1)

The equation of the line which passes through (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 ⇒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=0⇒ 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 + 3y = 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.

Q18.class 11 maths ex.miscellaneous

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)/2  + 3(8+y)/2 =7

x +3 +24 +3y =14

x + 3y + 13=0…….(i)

Since the line PQ ⊥ (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)

∴ (-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

Ans. The given equations are y = 3x+ 1….(i)  2y= x + 3 …..(ii) and  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) =±(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 +m²-6 -3m = 2m -1 +6m²-3m

5m² +5 = 0

m² +1 = 0

m= √-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 +m²-6 -3m = -2m +1 -6m²+3m

7m²-2m -7 =0

Hence the required value of m is (1±5√2)/7

Q20.If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y-5 = 0 and 3x – 2y + 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) are d1 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

√13x +√13 y -5√13 + 3√2 x -2√2 y +7√2 = 10√26

x(√13 +3√2) + y(√13 -2√2) +7√2-5√13 – 10√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 9x + 6y – 7 = 0 and 3x + 2y + 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 = ±3(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.

class 11 maths ex.miscellaneous ch.10 Q22

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

Drawing a normal BM⊥XX’

∠ABM =Incident angle=Φ, ∠CBM = Reflected angle=Φ(i.e reflected angle =incident angle)

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

∠OBA = 180°-(Φ+Φ+θ)= 180°-(2Φ+θ) = 180°-[2(90-θ)+θ)]= 180°-180 +2θ-θ =θ

∠ABX’ = 180°-θ

Therefore the slope of ray AB is = tan( 180°-θ)=-tanθ

The slope of BC is = tanθ

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

-tan θ =(0-2)/(a-1)

tan θ = 2/(a -1)……(i)

tan θ =(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 [√(a²-b²),0] and [-√(a²-b²),0] to the line (x/a)cosθ +(y/b) sinθ =1 is b²

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)cosθ +(y/b) sinθ =1

Wriiting it in the standard form

(bcosθ)x + (asinθ)y -ab =0

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

dand d2

Similarly

= b²

Hence the required product of the perpendiculars is b²

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

You can compensate us by donating any amount of money for our survival

Our Paytm NO 9891436286

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

NCERT Solutions of class 9 maths

Chapter 1- Number SystemChapter 9-Areas of parallelogram and triangles
Chapter 2-PolynomialChapter 10-Circles
Chapter 3- Coordinate GeometryChapter 11-Construction
Chapter 4- Linear equations in two variablesChapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s GeometryChapter 13-Surface Areas and Volumes
Chapter 6-Lines and AnglesChapter 14-Statistics
Chapter 7-TrianglesChapter 15-Probability
Chapter 8- Quadrilateral

NCERT Solutions of class 9 science 

Chapter 1-Matter in our surroundingsChapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?Chapter 10- Gravitation
Chapter3- Atoms and MoleculesChapter 11- Work and Energy
Chapter 4-Structure of the AtomChapter 12- Sound
Chapter 5-Fundamental unit of lifeChapter 13-Why do we fall ill ?
Chapter 6- TissuesChapter 14- Natural Resources
Chapter 7- Diversity in living organismChapter 15-Improvement in food resources
Chapter 8- MotionLast years question papers & sample papers

CBSE Class 9-Question paper of science 2020 with solutions

CBSE Class 9-Sample paper of science

CBSE Class 9-Unsolved question paper of science 2019

NCERT Solutions of class 10 maths

Chapter 1-Real numberChapter 9-Some application of Trigonometry
Chapter 2-PolynomialChapter 10-Circles
Chapter 3-Linear equationsChapter 11- Construction
Chapter 4- Quadratic equationsChapter 12-Area related to circle
Chapter 5-Arithmetic ProgressionChapter 13-Surface areas and Volume
Chapter 6-TriangleChapter 14-Statistics
Chapter 7- Co-ordinate geometryChapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

NCERT solutions of class 10 science

Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
Chapter 2- Acid, Base and SaltChapter 10- Light reflection and refraction
Chapter 3- Metals and Non-MetalsChapter 11- Human eye and colorful world
Chapter 4- Carbon and its CompoundsChapter 12- Electricity
Chapter 5-Periodic classification of elementsChapter 13-Magnetic effect of electric current
Chapter 6- Life ProcessChapter 14-Sources of Energy
Chapter 7-Control and CoordinationChapter 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-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbersChapter 13- Limits and Derivatives
Chapter 6- Linear InequalitiesChapter 14-Mathematical Reasoning
Chapter 7- Permutations and CombinationsChapter 15- Statistics
Chapter 8- Binomial Theorem 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 FunctionsChapter 9-Differential Equations
Chapter 2-Inverse Trigonometric FunctionsChapter 10-Vector Algebra
Chapter 3-MatricesChapter 11 – Three Dimensional Geometry
Chapter 4-DeterminantsChapter 12-Linear Programming
Chapter 5- Continuity and DifferentiabilityChapter 13-Probability
Chapter 6- Application of DerivationCBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Scroll to Top