NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.2) Exercise 10.2

The National Council of Educational Research and Training (NCERT) is a Government organization that was established to support and counsel the Central and State Governments on policies and programmes for the improvement of high-quality school education. The primary goals of NCERT and the units that make up the organization are to: conduct, promote, and coordinate research in fields related to school education; create and publish model textbooks, supplemental materials, newsletters, journals, and educational kits; and develop multimedia digital materials, among other things. It also organizes teacher pre-service and in-service training; encourages innovative educational techniques and practices; and collaborates and networks with state educational departments, universities, NGOs, and other educational institutions.

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.2) Exercise 10.2

Class 11 Maths Chapter 10 Exercise 10.2 explains how much distance is covered in a straight line by a point moving continuously in an uncurved direction. The various equations of the lines are Point-slope form, Two–point form, Slope-intercept form, Intercept-form, Normal form. Mathematics is termed as a critical subject as it consist of many equations and formulas. Students often face Mathematical problem. The Extramarks website provides NCERT Solutions For Class 11 Maths Chapter 10 Exercise 10.2. It provides solutions in a stepwise manner and in a clear way.

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

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Exercise 10.2

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NCERT Solutions for Class 11 Maths Chapters

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NCERT Solution Class 11 Maths of Chapter 10 All Exercises

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All Topics of Class 11 NCERT Maths for Exercise 10.2

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.2

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Q.1 In Exercises 1 to 8, find the equation of the line which satisfy the given conditions:

1. Write the equations for the x-and y-axes.
2.Passing through the point (

–4, 3) with slope 1 2 . MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabg gacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiA aiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabshacaqGOb GaaeyzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabcca caqGOaGaai4eGiaabsdacaqGSaGaaeiiaiaabodacaqGPaGaaeiiai aabEhacaqGPbGaaeiDaiaabIgacaqGGaGaae4CaiaabYgacaqGVbGa aeiCaiaabwgacaqGGaWaaSaaaeaacaqGXaaabaGaaeOmaaaacaqGUa aaaa@5E5B@

3. Passing through (0, 0) with slope m.
4.Passing through (2, 2

3 ) and inclined with the x-axis at an angle of 75°. MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr 4rNCHbGeaGqiVy0xg9vqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9 vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=x fr=xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaGaaeiuaiaabg gacaqGZbGaae4CaiaabMgacaqGUbGaae4zaiaabccacaqG0bGaaeiA aiaabkhacaqGVbGaaeyDaiaabEgacaqGObGaaeiiaiaabIcacaqGYa GaaeilaiaabccacaqGYaWaaOaaaeaacaqGZaaaleqaaOGaaeykaiaa bccacaqGHbGaaeOBaiaabsgacaqGGaGaaeyAaiaab6gacaqGJbGaae iBaiaabMgacaqGUbGaaeyzaiaabsgacaqGGaGaae4DaiaabMgacaqG 0bGaaeiAaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiEaiaab2 cacaqGHbGaaeiEaiaabMgacaqGZbGaaeiiaiaabggacaqG0bGaaeii aiaabggacaqGUbGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqGLb Gaaeiiaiaab+gacaqGMbGaaeiiaiaabEdacaqG1aGaaeiSaiaab6ca aaa@7313@

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2.
6. 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.
7. Passing through the points (–1, 1) and (2,– 4).
8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30°.

Ans.

1.
On x-axis,
Coordinate of y-axis = 0
So, the equation of x-axis:
y = 0
On y-axis,
Coordinate of x-axis = 0
So, the equation of y-axis:
x = 0

2.

Slope ​(m) of line=12          Given point=(−4,3)Equation of line having slope m and passing throughthe point(x1, y1)   y−y1=m(x−x1)     y−3=12(x+4)⇒         2(y−3)=(x+4)⇒             2y−6=x+4⇒    x−2y+10=0Thus, x−2y+10=0 is the required equation of the line.

3.

Slope ​(m) of line=m           Given point=(0,0)Equation of line having slope m and passing through the point(x1, y1)    y−y1=m(x−x1)      y−0=m(x−0)⇒            y=mxThus, y=mx is the required equation of the line.

4.

Slope ​(m) of line=tan75°       =tan(45°+30°)       =tan45°+tan30°1−tan45°tan30°       =1+131−1.13       =3+13−1          Given point=(2,23)

Equation of line having slope m and passing through the point(x1, y1)⇒                   y−y1=m(x−x1)⇒                y−23=3+13−1(x−2)⇒                      (y−23)(3−1)=(3+1)(x−2)⇒      y(3−1)−23(3−1)=(3+1)x−2(3+1)⇒ (3+1)x−y(3−1)+23(3−1)−2(3+1)=0⇒        (3+1)x−(3−1)y+6−23−23−2=0⇒                 (3+1)x−(3−1)y+4−43=0⇒             (3+1)x−(3−1)y+4(1−3)=0⇒                     (3+1)x−(3−1)y=4(3−1)Thus, it is the required equation of the line.

5.

The coordinate of the point lying on x-axis at a distance of 3 units to the left of the origin=(−3,0)          Slope of the line=−2Eqation of the line having slope −2 and passing through the point (−3,0) is given as:   

  y−0=−2(x+3)      [∵y−y1=m(x−x1)]⇒     y=−2x−6⇒    2x+y+6=0Which is the required equation of the given line. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@

6.

The distance on y-axis of intersection point by line,(c)=2  Angle​ made by line with x-axis=30°  Slope of given line,m=tan30° =13Equation of line having slope 13 and having intercept on y-axisis given as: y=13x+2[∵y=mx+c]⇒    3y=x+23⇒    x−3y+23=0Which is the required equation of the given line.

7.

Equation of line passing through two point (x1,y1) and (x2,y2) is given as: y−y1=y2−y1x2−x1(x−x1)

Equation of line passing through two point (−1, 1) and (2,− 4)is:   y−1=− 4−12+1(x+1)⇒  y−1=− 53(x+1)⇒      3(y−1)=− 5x−5⇒       5x+3y+2=0Which is the required equation of the line.

8.

Hence, the equation of the line having normal distance p from the origin and angle ω which the normal makes with the positive direction of x-axis is given byx cos ω + y sin ω=pHere, ω=30°​ and p=5 unitsSo, equation of the line is:  x cos 30° + y sin 30°=5⇒       x(32) + y(12)=5⇒ 3x+y=10

Which​​ is the required solution of the given equation.

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

Ans.

Coordinates of vertices of ΔPQR are:P(2, 1), Q(–2, 3) and R(4, 5)Median through point R bisects side PQ of Δ PQR.Mid-point of PQ={2+(−2)2,1+32}      =(0,2)Equation of median passing through (4,5) and (0,2) is:  y−y1=y2−y1x2−x1(x−x1)⇒     y−5=2−50−4(x−4)⇒     y−5=−3−4(x−4)⇒          4y−20=3x−123x−4y+20−12=0⇒        3x−4y+8=0Therefore, it is the equation of median passing through vertex R.

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

Ans.

Slope of the line through the points (2,5) and  (−3,  6)=6−5−3−2         =1−5Slope of the⊥ on the line=−1(1−5)         =5Equation​ of the perpendicular on the line passing through the point (−3,5) is:       y−5=5(x+3)⇒      y−5=5x+15⇒       5x+15−y+5=0⇒            5x−y+20=0Which is the required equation of line.

Q.4 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.

Ans.

Slope​ of the line segment joining the points (1,0) and(2,3)=3−02−1=3Slope​ of perpendicular to the line segment joining the points

(1,0) and (2,3)=−13Coordinate​ of the point dividing the line segment joining the points (1,0) and (2,3) in the ratio of 1:n is:(1.2+n.11+n,1.3+n.01+n)=(2+n1+n,31+n)Equation of line intersecting line segment in the ratio of 1:n is:           y−31+n=−13(x−2+n1+n)⇒              (1+n)y−31+n=−13{(1+n)x−2−n1+n}⇒           3(1+n)y−9=−(1+n)x+2+n⇒          (1+n)x+3(1+n)y−2−n−9=0⇒                     (1+n)x+3(1+n)y=n+11Which is the required equation of the line.

Q.5 Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Ans.

Equation​ of line in the form of intercepts on axis is: xa+yb=1Since, a=b. Then, xa+ya=1 ...(i) Since, this line passes through the point (2,3).So, 2a+3a=1⇒        5a=1⇒         a=5Putting value of a in equation (i),​ we get x5+y5=1⇒  x+y=5Which is the required equation of line.

Q.6 Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.

Ans.

Equation​ of line in the form of intercepts on axis is: xa+yb=1Since, a+b=9. Then putting b=9−a, we get     xa+y9−a=1 ...(i)Since, this line passes through the point (2,2).So,    2a+29−a=1⇒    2(9−a)+2a(9−a)a=1

⇒           18=9a−a2⇒     a2−9a+18=0⇒   (a−6)(a−3)=0⇒         a=6,3Since,        b=9−aSo,       b=3,6Thus, the equation of line is           x6+y3=1⇒x+2y=6or           x3+y6=1⇒2x+y=6

Q.7

Find equation of the line through the point (0, 2) making an angle 2π3 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.

Ans.

Equation of line through the point (0,2) making an angle 2π3 with the positive x-axis is y−2=tan(2π3)(x−0)⇒ y−2=tan(π−π3)x

⇒ y−2=−tan(π3)x⇒ y−2=− 3 x⇒ 3x+y−2=0Slope of parallel line to 3x+y−2=0, is        m=− 3Equation of line of slope − 3 and y-intercept −2 is:        y=−  3x−2[y=mx+C]        3x+y+2=0Therefore, the required equations are: 3x+y−2=0 and 3x+y+2=0.

Q.8 The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

Ans.

Slope of perpendicular passing through 0,0 and –2,9 = 9–0–2–0                                                                             = –92Slope of perpendicular line= –1–92                                       =29Equation of line passing through –2,9 and having slope 29 is               y–9=29x+2           9y–81=2x+4⇒ 2x–9y+85=0Which is required equation of line.

Q.9 The length L (in centimetre) 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.

Ans.

Since, copper rod is a linear function of its Celsius temperature C. In an experiment L=124.942 when C=20 and L=125.134 when C=110Let two points which satisfy the linear function between L and C are (124.942,20) and (125.134,110).Let us consider that L is along x-axis and C is along y-axis.So, the equation of line passing through the points (124.942,20) and (125.134,110) in XY-plane. L−124.942 =125.134−124.942110−20(C−20)⇒L−124.942 =0.19290(C−20)⇒ L=0.19290(C−20)+124.942

Which is the required linear function of length a copper rod.

Q.10 The owner of a milk store finds that, he can sell 980 litres of milk each week at ₹14/litre and 1220 litres of milk each week at ₹16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at ₹17/litre?

Ans.

Let selling price per litre is along x-axis and demand along y–axis. Two points 14,980 and 16,1220 are in XY–planeand they satisfy the linear relation between selling price anddemand of milk. Then,y–980=1220–98016–14x–14⇒         y–980=2402x–14⇒           y=120x–14+980   ...iWhen x=₹17/litre, from equation i:⇒         y=12017–14+980⇒         y=360+980    =1340 litresThus, the owner of a milk store can sell 1340 litres of milk each week at ₹17/litre.

Q.11

P(a, b) is the mid-point of a line segment between axes.Show that equation of the line is  xa+yb=1.

Ans.

Let coordinates of A and B are 0,y and x,0 respectively.Since, Pa,b is mid point of AB.So,a=0+x2 and b=y+02⇒      x=2a and  y=2bi.e., OB=2a​ and OA=2bEquation of line in the intercept form is     xOB+yOA=1⇒        x2a+y2b=1⇒             xa+yb=2which is the required equation.

Q.12 Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Ans.

Let coordinates of P and Q are (0,y) and (x,0) respectively.Since, R(h,k) divides PQ in ratio 1:2, i.e., QR : QP=1:2. So,      h=2.x+1.01+2 and k=2.0+1.y1+2⇒          x=32 h and    y=3ki.e., OQ=32 h​ and OP=3kEquation of line in the intercept form is      xOQ+yOP=1⇒ x32 h​+y3k=1⇒        2xh+yk=3

⇒        2kx+hy=3hkTherefore, the required equation is 2kx+hy=3hk.

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