NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.1) Exercise 10.1

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines (Ex 10.1) Exercise 10.1

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Class 11 Maths Chapter 10 Exercise 10.1 explains that a point travelling steadily in a direction with no curvature covers a certain distance in a straight line. The fundamental concepts of lines, include slopes, angles between two lines, various sorts of lines, and the separation between lines. Extramarks is a website that contains Class 11 Maths NCERT Solutions Chapter 10 Exercise 10.1, students can understand all the concepts at Extramarks.

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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

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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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NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Exercise 10.1

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Q.1 Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area.

Ans.

The given points are:A(−4,5),B(0,7),C(5,−5) and D(− 4,−2).Area of ΔABC=|12{x1(y2−y3)+x2(y3−y1)+x3(y1−y2)}|=|12{−4(7+5)+0(−5+5)+5(5−7)}|=|12(− 48+0−10)|=29 sq. units

Area of ΔADC=|12{−4(−2+5)−4(−5−5)+5(5+2)}|=|12(− 12+40+35)|=632 sq. units Area (ABCD)=Area of ΔABC+Area of ΔADC=(29+632) sq. units=1212 sq. unitsThus, the area of quadrilateral is 1212  square units.

Q.2 The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Ans.

Coordinate of B and C are (0,– a) and (0,a) respectively.In​ ΔAOB, ∠AOB=90°So,​ by Pythagoras theorem AB2=AO2+OB2⇒   (2a)2=AO2+a2⇒      AO=4a2−a2         =±3aSo, the coordinate of A is (0,3a) or (0,−3a).Therefore,​ the coordinates of triangle are:(0,a), (0,− a)  and  (0,3a)Or  (0,a), (0,− a) and  (0,− 3a)

Q.3 Find the distance between P (x1, y1) and Q (x2, y2) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.

Ans.

Distance between two points P(x1,y1) and Q(x2,y2) =|(x2−x1)2+(y2−y1)2|

(i) When PQ is parallel to y-axis: Then, x1=x2PQ=|(x1−x1)2+(y2−y1)2|PQ=|(y2−y1)2|PQ=|(y2−y1)|

(ii) When PQ is parallel to x-axis: Then, y1=y2PQ=|(x2−x1)2+(y2−y1)2|PQ=|(x2−x1)2|PQ=|(x2−x1)|

Q.4 Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Ans.

Let the point on x-axis be P(a,0).Given points are A(7,6) and B(3,4).Then, according to given condition:          PA=PB⇒|(a−7)2+(0−6)2|=|(a−3)2+(0−4)2|⇒       |(a−7)2+36|=|(a−3)2+16|⇒     a2−14a+49+36=a2−6a+9+16⇒       49+36−25=14a−6a⇒       49+36−25=8a⇒                 60=8a⇒                    a=608⇒                    a=152

Therefore, the point on x-axis is ( 15 2 ,0 ). MathType@MTEF@5@5@+=feaaguart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaacaWGubGaamiAaiaadwgacaWGYbGaamyzaiaadAgacaWGVbGaamOCaiaadwgacaGGSaGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabccacaqGVbGaaeOBaiaabccacaqG4bGaaeylaiaabggacaqG4bGaaeyAaiaabohacaqGGaGaaeyAaiaabohacaqGGaWaaeWaaeaadaWcaaqaaiaaigdacaaI1aaabaGaaGOmaaaacaGGSaGaaGimaaGaayjkaiaawMcaaiaac6caaaa@5C7F@

Q.5 Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Ans.

Mid−point of P (0,− 4) and B (8, 0)=(0+82,− 4+02)=(4,−2)Slope of line passing through origin(0,0) and (4,− 2)=−2−04−0=−24=−12Therefore, the required slope of line is −12.

Q.6 Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.

Ans.

Given points are: (4, 4), (3, 5) and (–1,–1)Slope of (4, 4)  and  (3, 5), m1=5−43−4                   =1−1                    =−1

Slope of (3, 5)  and  (–1,–1), m2=−1−4−1−3                  =−5−4                  =54Slope of (–1,–1)  and  (4, 4), m3=4+14+1                  =55                  =1Here, m1.m3=−1×1                     =−1So, the sides of triangle having slope m1 and m3 are perpendicular to each other.Therefore, given points are the vertiaces of a right angled triangle.

Q.7 Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Ans.

Angle of line with y-axis is 30°.
Angle of line with x-axis = 90° + 30° = 150°
Slope of line = tan150° = – √3

Q.8 Find the value of x for which the points (x, – 1), (2, 1) and (4, 5) are collinear.

Ans.

Let points are A(x,−1), B(2,1) and C(4, 5).Slope of AB=1+12−x        =22−xSlope of BC=5−14−2        =42        =2Slope of CA=−1−54−xIf A, B and C are collinear.Slope of AB=Slope of BCSo,       22−x=2⇒           2−x=1

⇒                  x=1 For x=1, point A, B and C are collinear. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiabgkDiElaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaadIhacqGH9aqpcaaIXaaabaGaamOraiaad+gacaWGYbGaaeiiaiaabIhacqGH9aqpcaqGXaGaaeilaiaabccacaqGWbGaae4BaiaabMgacaqGUbGaaeiDaiaabccacaqGbbGaaeilaiaabccacaqGcbGaaeiiaiaabggacaqGUbGaaeizaiaabccacaqGdbGaaeiiaiaabggacaqGYbGaaeyzaiaabccacaqGJbGaae4BaiaabYgacaqGSbGaaeyAaiaab6gacaqGLbGaaeyyaiaabkhacaqGUaaaaaa@7B5F@

Q.9 Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Ans.

The given points are let A−2,−1,B4, 0,C3, 3 and D −3, 2.Slope of AB=0+14+2                   =16Slope of BC=3−03−4                  =3−1                  =−3Slope of CD=2−3−3−3                 =−1−6                 =16Slope of DA=−1−2−2+3                  =−31                  =−3Since, the slope of AB=the slope of CDSo, AB is parallel to CD.And, the slope of BC=the slope of ADSo, BC is parallel to AD.Therefore, ABCD is a parallelogram.

Q.10 Find the angle between the x–axis and the line joining the points (3,–1) and (4,–2).

Ans.

m=Slope of 3,–1 and 4,–2 = –2+14–3   = –11   = –1Let line makes θ angle with x-axis. Then, tanθ = –1   =tan135°⇒  θ=135°Thus, the angle between x-axis and line joining the given point is 135°.

Q.11

The slope of a line is double of the slope of another line. If tangent of the angle between them is 13, find the slopes of the lines.

Ans.
Let the slope of first line, m1 = 2m
Slope of other line, m2 = m
Let angle between two lines be θ.

tanθ=|m2−m11+m1m2|⇒            13=|m−2m1+2m.m|⇒            13=|−m1+2m2|⇒            13=−m1+2m2 or  13=−(−m1+2m2)⇒            13=−m1+2m2   or        13=m1+2m2⇒ 2m2+1=−3m     or   2m2+1=3m⇒2m2+3m+1=0     or   2m2−3m+1=0⇒        2m2+2m+m+1=0or2m2−2m−m+1=0⇒ 2m(m+1)+1(m+1)=0or2m(m−1)−1(m−1)=0⇒            (m+1)(2m+1)=0or(m−1)(2m−1)=0⇒ m=−1,−12orm=−1,−12∴m1=2(−1)   or m1=2(−12)  or    m1=2(1)or m1=2(12)     m1=−2,−1,2,1 and m2=−1,−12,−1,−12Thus, the slope of the lines are 1 and 2,12 and 1,−1 and−2 or − 12 and−1.

Q.12 A line passes through (x1, y1) and (h, k). If slope of the line is m, show that k – y1 = m (h – x1).

Ans.

Slope of line passing through (x1,y1) and (h,k)=k−y1h−x1But, the slope of line=mthen,         m=k−y1h−x1⇒ k−y1=m(h−x1).

Q.13 

If three points (h, 0), (a, b) and (0, k) lie on a line, show that ah+bk=1.

Ans.

Let the threepointsA(h,0),B(a,b)andC(0,k) lie on a line.Then, slope of AB=slope of BC⇒         b−0a−h=k−b0−a⇒           −ab=(k−b)(a−h)⇒          − ab=ak−kh−ab+bh⇒                 0=ak−kh+bh⇒     ak+bh=khDividing both sides by kh, we get   akkh+bhkh=khkh

⇒     ah+bk=1Hence proved.

Q.14 Consider the following population and year graph (Fig 10.10), find the slope of the line AB and using it, find what will be the population in the year 2010?

Ans.

Let the coordinates of point A and B are ( 1985, 92 ) and ( 1995, 97 ) respectively. Then, slope of line AB= 97−92 1995−1985 = 5 10 = 1 2 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AB7C@

Let population in the year 2010=kThen, coordinate of point C lying on line AB=(2010,k)Slope of line BC=k−972010−1995=k−9725Since, A, B and C lies on the same line. So,Slope of line AB=Slope of line BC⇒    12=k−9725⇒  25=2k−194⇒       25 +194=2k⇒                   219=2k⇒    k=2192        =104.5Thus, slope of line is 12 and population in 2010 was 104.5 crores.

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