NCERT Solutions For Class 11 chapter-10 Straight Lines Exercise 10.2

NCERT Solutions for Class-11 Maths Chapter-10 Straight LInes

NCERT Solutions For Class 11 Maths Chapter-10 Straight LInes Exercise 10.2 prepared by the expert of Physics Wallah score more with Physics Wallah NCERT Class 11 maths solutions. You can download solution of all chapters from Physics Wallah NCERT Solutions of class 11.

 

NCERT Solutions for Class-11 Maths Exercise 10.2

 

In Exercises 1 to 8, find the equations of the line which satisfy the given conditions:

 

Question1. Write the equations for the x-and y-axes.

Solution :
The y-coordinate of every point on the x-axis is 0.

Therefore, the equation of the x-axis is y = 0.

The x-coordinate of every point on the y-axis is 0.

Therefore, the equation of the y-axis is x = 0. 

Question2. Passing through the point (– 4, 3) with slope 1/2

Solution :Given that, the point (−4,3) and slope of the line is 1 / 2.

The equation of the line passing through point (x0,y0), whose slope is m, is

(y−y0)=m(x−x0).

Thus, the equation of the line passing through point (−4,3), whose slope is 1 / 2, is (y−3) = 1 / 2 (x+4)

Cross multiply,

2(y−3)=x+4

2y−6=x+4

Move the constants together,
x−2y+10=0

Therefore, the equation of the line which passes through the point (−4,3)
 with slope 12 is x−2y+10=0.

Question3. Passing through (0, 0) with slope m.

Solution :Given that, the point is  (0,0) and slope is m.

The equation of the line passing through point (x0,y0) , whose slope is m, is

(y−y₀) = m(x−x₀).
Substitute the values in the equation,

⇒(y−0) = m(x−0)

⇒y = mx

Therefore, the equation of the line which passes through (0,0) with slope m
is y = mx.

Question4. Passing through (2,2√3) and inclined with the x-axis at an angle of 75^∘

Solution :Given that, the point is  (2,2√3)

The slope of the line that inclines with the x-axis at an angle of 75^∘

That is, m = tan75^∘

m = tan (45^∘ + 45^∘)

The equation of the line passing through point (x⁰,y⁰), whose slope is m, is (y−y₀) = m(x−x₀).
Thus, if a line passes through (2,2√3) and inclines with the x-axis at an angle of 75^∘,then the equation of the line is,
 

Therefore, the equation of the line which passes through (2,2√3) and is inclined with the x-axis at an angle of 75∘is (√3+1)x−(√3−1)y = 4(√3−1).

Question5: Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.

Solution :
It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as

y = m(x – d)

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.

The slope of the line is given as m = –2

Thus, the required equation of the given line is

y = –2 [x – (–3)]

y = –2x – 6

i.e., 2x + y + 6 = 0 

Question6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30^∘with positive direction of the x-axis.

Solution :Given, the line intersects the y-axis at a distance of 2
units above the Origin.The line makes an angle of 30∘ with the positive direction of the x-axis.
That is, c=2

m=tan30^∘

=–√13

If a line with slope  makes y - intercept c, then the equation of the line is,

y=mx+c

Substitute the values,

Therefore, the equation of the line which intersects the y-axis at a distance of 2
 units above the origin and makes an angle of 30^∘ with the positive direction of the x-axis is x−√3y+ √23 =0

Question7. Passing through the points (–1, 1) and (2, – 4).

Solution :
 

Given that the line makes angle 30^∘ with positive direction of y- axis.

Thus, the angle made by line with x axis is,

90+30 = 120∘

Slope m = tanθ

=tan120^∘

=−tan60∘

= √3

Therefore, the slope of the line, which makes an angle of 30∘
 with the positive direction of y-axis measured anticlockwise is −√3

Question8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30^∘.

Solution :.Let p be the length of the normal from the origin to a line and ω be the angle made by the normal with the positive direction of the x-axis

Now, equation of the line is xcosω+ysinω=p

Here, p=5 units and ω=30^∘

 
Thus, the equation of the given line is,

xcos30^∘ + ysin30^∘ =5

Substitute the values,

x√3 / 2 + y⋅1 / 2 = 5

That is, √3 x + y = 10

Therefore, the equation of the line which is at a perpendicular distance of 5
 units from the origin and the angle made by the perpendicular with the positive x-axis is 30∘is √3x + y = 10.

Question9. 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 :Given that, the vertices of ΔPQR are P(2,1),Q(−2,3) and R(4,5)

Let RL
 be the median through vertex R

And L be the mid-point of PQ

By mid-point formula, the coordinates of point L
 are,

Equation of required median RS is

Question10. Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

Solution :The slope of the line joining the points (2,5) and (−3,6)

 is,

m= 6−5 / −3−2

= 1 / −5

 

It is known that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Thus, slope of the line perpendicular to the line through the points (2,5)
 and (−3,6)

 is,



The equation of the line passing through point (−3,5)
, whose slope is5

is,

 (y−5) = 5(x+3)

Expand brackets,

y−5 = 5x+15

That is, 5x−y+20=0

Therefore, the equation of the line passing through (−3,5)
 and perpendicular to the line through the points (2,5) and (−3,6) is 5x−y+20=0.

Question11. 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 :By section formula, the coordinates of the point that divides the line segment joining the points (1,0) and (2,3) in the ration 1:n

 is,




It is known that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Thus, slope of the line that is perpendicular to the line joining the points (1,0)
 and (2,3)

 is,



The equation of the line passing through and whose slope is −1 / 3

 is ,

Therefore, equation of the line perpendicular to the line segment joining the points (1,0)
 and (2,3) divides it in the ration 1:n is (1+n)x+3(1+n)y=n+11.

 

Question12. Find the equation of the line cuts off equal intercepts on the coordinate axis and passes through the point (2, 3).

Solution :The equation of a line in the intercept form is xa + yb =1

Here,  a andb
are the intercepts on x and y axes respectively.

Given that,  the line cuts off equal intercepts on both the axes.

That is, a=b

.

Now, xa + ya = 1

⇒ x + y = a

The given line passes through the point (2,3)

, this equation reduces to

 2 + 3 = a

 ⇒a=5

 

Substitute the value of a
 in x + y =  a

That is, x + y = 5

Therefore, the equation of a line that cuts off equal intercepts on the coordinate axes

and passes through the points (2,3)
 is x + y = 5.

 

Question13. Find the equation of the line passing through the point (2, 2) and cutting off intercept on the axis whose sum is 9.

Solution :
Given: Line passes through point (2, 2).  a = 6 , and b = 9-6 = 3 then the equation of the line is 

x/ 6 + y/3 = 1 ⇒ x + 2y - 6 = 0

if a = 3 and b = 9-3  = 6  , then the equation  of the line is

x/3 + y/6 = 1 ⇒ 2x + y -6 = 0

the equatiuon  of a line in the intercept form is  x/a +y/b = 1  --- (1)

Question14. Find equation of the line through the point (0, 2) making an angle 2π/3 with the positive x-axis. 

Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

Solution :
 

 

Question15. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Solution :Given that, The perpendicular from the origin to a line meets it at the point (−2,9)

.

The slope of the line joining the origin (0,0)
 and point (−2,9)

 is,


Then, the slope of the line perpendicular to the line joining the origin and points (−2,9)

 is,



Now, the equation of the line passing through point (−2,9)
 and having a slope m2 is,(y−9)=2/ 9(x+2)

Cross multiply and expand brackets,

9y−81 = 2x+4

That is, 2x−9y+85=0

Therefore, the equation of the line is 2x−9y+85=0
.
 

Question16. The length L (in centimeter) of a copper rod is a linear function of its Celsius temperature C. In an experiment if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.

Solution :It is given that when C=20,  L=124.942 and when C=110,  L=125.134

The points (20,124.942)
 and (110,125.134) satisfy the linear relation between Land C.

Assume C
along the x
-axis and L along the y
-axis, we have two points, (20,124.942)

 

and (110,125.134)
 in the XY

 plane.

Thus, the linear relation between L
 and C

 is the equation of the line passing through

thepoints (20,124.342)
 and (110,125.134)

Question17. The owner of a milk store finds that he can sell 980 liters of milk each week at Rs. 14 litre and 1220 liters of milk each week at Rs. 16 liters. Assuming a linear relationship between selling price and demand, how many liters could he sell weekly at Rs. 17 liter?

Solution :Given that, the owner can sell 980
 liters of milk each week at Rs 14/ liter and 1220
 liters of milk each week at Rs 16/

 liter.

The relationship between the selling price and demand is linear.

Assume selling price per liter along the x
-axis and demand along the y
-axis, we have two points (14,980) and (16,1220) in the XY

 plane that satisfy the linear relationship between selling price and demand.

Thus, the line passing through points (14,980)
 and (16,1220)

That is, y−980=((1220−980) / (16−14)) (x−14)

y−980=(240 / 2) (x−14)

y−980=120(x−14)

y=120(x−14)+980

If x= Rs 17 / liter,

y=120(17−14) + 980

⇒y = 120×3 + 980

 =360 + 980

 =1340

Therefore, the owner of the milk store could sell 1340
 liters of milk weekly at Rs 17/liter

Question18. P(a , b) is the mid-point of a line segment between axis. Show that equation of the line is

Solution :
 

Then

 

 

 

Question19. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find bequation of the line.

Solution :
Let AB and BA be two points where the line intersect p (a,b)and axis respectively and R is a point divides AB in the ratio 1: 2.

 

Question20. By using the concept of equation of a line, prove that the three points (3, 0),(– 2, – 2) and (8, 2) are collinear.

Solution :It is necessary to show that the line passing through points (3,0) and (−2,−2) also passes through point (8,2)in order to show that the points (3,0),(−2,−2) and (8,2)

 are collinear
The equation of the line passing through points (3,0)
 and (−2,−2) is,

.Thus, the line passing through points (3,0) and (−2,−2) also passes through point (8,2)

.Therefore, the points (3,0),(−2,−2)
, and (8,2) are collinear.