Class 11 Maths Ncert Solutions Chapter 10 Ex 10.1 Straight Lines PDF

# Class 11 Maths Ncert Solutions Ex 10.1

## Class 11 Maths Ncert Solutions Chapter 10 Ex 10.1

Q-1. Construct a quadrilateral in the Cartesian plane with vertices (-2, 5), (0, 6), (4, -4) and (-3, -1). Also, find the area of the quadrilateral.

Solution.

Let, MNOP be the given quadrilateral having vertices M (-2, 5), N (0, 6), O (4, -4) and P (-3, -1).

Now, plot M, N, O and P on the Cartesian plane and join MN, NO, OP and PM.

The quadrilateral, thus, formed will be

As we need to find the area of the quadrilateral MNOP, so draw a diagonal, say, MO.

Now,

Area of (MNOP) = area of (ΔMNO$\Delta MNO$) + area of ( ΔMOP$\Delta MOP$ )

Area of a triangle with vertices (a1, b1), (a2, b2), (a3, b3) and (a4, b4) is given by:

12$\frac{ 1 }{ 2 }$ | a1 (b2 – b3 ) + a2 (b3 – b1 ) + a3 (b1 – b2 ) |

Now,

Vertices of ΔMNO$\Delta MNO$ is M (-2, 5), N (0, 6) and O (4, -4)

Area of ΔMNO$\Delta MNO$ = 12|[2(6(4))+0(4(5))+4(5(6))]|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -6 – \left ( -4 \right ) \right ) + 0\left ( -4 – \left ( 5\right ) \right ) + 4\left ( 5 – \left ( 6 \right ) \right ) \right ] \right |$

= 12|[2(6+4)+0(45)+4(56)]|=12|[20+04]|=12|24|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -6 + 4 \right ) + 0\left ( -4 – 5 \right ) + 4\left ( 5 – 6 \right ) \right ] \right | \\ = \frac{ 1 }{ 2 }\left | \left [ -20 + 0 – 4 \right ] \right | \\ = \frac{ 1 }{ 2 }\left | -24 \right |$

= 12×24$\frac{ 1 }{ 2 } \times 24$

= 12 unit2

Vertices of ΔMOP$\Delta MOP$ is M (-2, 5), O (4, -4) and P (-3, -1).

Area of ΔMOP$\Delta MOP$ = 12|[2(4(1))+4(1(5))3(5(4))]|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -4 – \left ( -1 \right ) \right ) + 4\left ( -1 – \left ( 5\right ) \right ) – 3\left ( 5 – \left ( -4 \right ) \right ) \right ] \right |$

= 12|[2(4+1)+4(15)3(5+4)]|=12|[62427]|=12|45|$\frac{ 1 }{ 2 }\left | \left [ -2\left ( -4 + 1 \right ) + 4\left ( -1 – 5 \right ) – 3\left ( 5 + 4 \right ) \right ] \right | \\ = \frac{ 1 }{ 2 }\left | \left [ 6 – 24 – 27 \right ] \right | \\ = \frac{ 1 }{ 2 }\left | -45 \right |$

= 12×45$\frac{ 1 }{ 2 } \times 45$

= 22.5 unit2

Area of MNOP = Area of ΔMNO$\Delta MNO$ + Area of ΔMOP$\Delta MOP$

= 12 + 22.5 = 34.5 22.5 unit2

Q-2. Consider an equilateral triangle, each of whose sides are 2b which lies on the y- axis in such a manner that, the mid- point of each of its base is at the origin. Obtain all the vertices of the equilateral triangle.

Solution.

Let, XYZ be the given equilateral triangle each of whose side is 2b.

Since, XYZ is an equilateral triangle,

So, XY = YZ = ZX = 2b

Now, consider that the base YZ lies along y-axis in such a way that it’s mid- point is at the origin.

So, YA = AZ = b, where A is the origin.

Thus, by our observation, the coordinates of the point Y are (0, b), while the coordinates of the point Z is (0, -a).

We know that, the line joining the vertex of an equilateral triangle having the mid- point of its opposite sides is perpendicular.

Thus, the vertex X lies on the y- axis.

By applying the Pythagoras theorem to ΔXAZ$\Delta XAZ$, we will get

( XZ )2 = ( AX )2 + ( AZ )2

(2b)2=(AX)2+(b)2$\Rightarrow \left ( 2b \right )^{ 2 } = \left ( AX \right )^{ 2 } + \left ( b \right )^{ 2 }$ (AX)2=(4b2b2)$\Rightarrow \left ( AX \right )^{ 2 } = \left ( 4b^{ 2 } – b^{ 2 } \right )$ (AX)2=3b2$\Rightarrow \left ( AX \right )^{ 2 } = 3b^{ 2 }$ AX=±3b$\Rightarrow AX = \pm \sqrt{ 3 }b$

The coordinate of the point X = ( ±3b$\pm \sqrt{ 3 }b$ , 0 )

Hence, the vertices of the equilateral triangle XYZ are X( 3b$\sqrt{ 3 }b$ , 0 ), Y( 0, b ) and Z( 0, -b ) or X( 3b$-\sqrt{ 3 }b$ , 0 ), Y( 0, b ) and Z( 0, -b )

Q-3. What is the distance between X (a1, b1) and Y (a2, b2) if:

(a) XY is parallel to x- axis

(b) XY is parallel to y- axis

Solution.

Given points are X (a1, b1) and Y (a2, b2)

(a) The given condition is:

XY is parallel to the x- axis, i.e., b1 = b2

Therefore, the distance hence measured between X and Y in this case = (a2a1)2+(b2b1)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= (a2a1)2+(b2b2)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 2 } \right )^{ 2 } }$

= (a2a1)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } }$

= | a2 – a1 |

(b) The given condition is:

XY is parallel to the y- axis, i.e., a1 = a2

Therefore, the distance hence measured between X and Y in this case = (a2a1)2+(b2b1)2$\sqrt{ \left ( a_{ 2 } – a_{ 1 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= (a2a2)2+(b2b1)2$\sqrt{ \left ( a_{ 2 } – a_{ 2 } \right )^{ 2 } + \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= (b2b1)2$\sqrt{ \left ( b_{ 2 } – b_{ 1 } \right )^{ 2 } }$

= | b2 – b1 |

Q-4. Consider two points (6, 5) and (4, 2). Get a point on the y- axis which is equivalent from the given two points.

Solution.

Let us assume that, (0, y) be the given point on y- axis which is equivalent to the two given points (6, 5) and (4, 2).

Now,

(60)2+(5a)2=(40)2+(2a)2$\sqrt{ \left ( 6 – 0 \right )^{ 2 } + \left ( 5 – a \right )^{ 2 }} = \sqrt{ \left ( 4 – 0 \right )^{ 2 } + \left ( 2 – a \right )^{ 2 }}$ 36+2510a+a2=16+44a+a2$\Rightarrow \sqrt{ 36 + 25 – 10a + a^{ 2 }} = \Rightarrow \sqrt{ 16 + 4 – 4a + a^{ 2 }}$ 41=6a$\Rightarrow \sqrt{ 41 } = \sqrt{ 6a }$

Squaring both side, we will get

6a = 41

a = 416$\frac{ 41 }{ 6 }$

a = 6.8 unit

Therefore, (0, 416$\frac{ 41 }{ 6 }$ ) is the point on Y – axis which is equivalent to the given points (6, 5) and (4, 2).

Q-5. What is the slope of a line passing through the origin and, the mid- point of the line- segment joining the two points O (0, -5) and A (9, 0)?

Solution.

The two points of the line- segment is O (0, -5) and A (9, 0).

Then, the co-ordinates of the mid-point of line-segment OA are-

(0+92,5+02)$\left( \frac{ 0 + 9 }{ 2 } , \frac{ -5 + 0 }{ 2 } \right )$ = ( 4.5, -2.5 )

We know that the slope (say, m) of the non- vertical line which passes through the points (a1, b1) and (a2, b2) is:

m = b1b2a1a2$\frac{ b_{1} – b_{2} }{ a_{1} – a_{2} }$, a2 ≠ a1

Hence, the slope of the line- segment passing through (0, 0) and (4.5, -2.5) is given by:

2.504.50=2.54.5=2545=59$\frac{ -2.5 – 0 }{ 4.5 – 0 } = \frac{ -2.5 }{ 4.5 } = \frac{ -25 }{ 45 } = \frac{ -5 }{ 9 }$

Therefore, the required slope of the line-segment = 59$\frac{ -5 }{ 9 }$

Q-6. Prove that the points (5, 5), (4, 6) and (-2, -2) are the vertices of the right- angled triangle, without using Pythagoras theorem.

Solution.

There are three points given in the question. Let, those points be the vertices of the triangle XYZ, namely, X (5, 5), Y (4, 6) and Z (-2, -2).

We know that, the slope (say, m) of the non- vertical line which passes through the points (a1, b1) and (a2, b2) is:

m = b1b2a1a2$\frac{ b_{1} – b_{2} }{ a_{1} – a_{2} }$, a2 ≠ a1

The slope of XY( m1 ) = 6545$\frac{ 6 – 5 }{ 4 – 5 }$ = -1.

The slope of YZ( m2 ) = 2624=86=43$\frac{ -2 – 6 }{ -2 – 4} = \frac{ -8 }{ -6 } = \frac{ 4 }{ 3 }$

The slope of ZX( m3 ) = 5+25+2=77=1$\frac{ 5 + 2 }{ 5 + 2 } = \frac{ 7 }{ 7 } = 1$

Here, we can see that, m1. m3 = -1

This proves that, the line- segments XY and ZX is perpendicular to each other, i.e., the triangle XYZ is right- angled at X (5, 5).

Hence, proved that the given points, i.e., (5, 5), (4, 6) and (-2, -2) are the vertices of the right- angled triangle.

Q-7. What is the slope of the line which makes an angle 60$60^{\circ}$ along the positive direction of the Y- axis which is measured in anticlockwise sequence.

Solution.

By mathematical calculation rule,

If a line makes an angle of 60$60^{\circ}$ with any of the positive direction of the axis which is measured in an anticlockwise sequence, then the angle made by the line with positive direction of the corresponding axis is also measured in anticlockwise sequence is given by,

90+x$90^{\circ} + x^{\circ}$.

Then, if the angle measured by the line in the x- axis measured in positive direction of the x- axis, measured in anticlockwise direction is-

90+60=150$90^{\circ} + 60^{\circ} = 150^{\circ}$

Therefore, the slope of the given line is tan 150$150^{\circ}$ = tan( 18030)$180^{\circ} – 30^{\circ} )$ = – tan 30$30^{\circ}$ = 13$\frac{ 1 }{ \sqrt{ 3 }}$.

Q-8. What will be the value of a so that the points (a, -2), (3, 2) and (5, 6) get collinear to each other?

Solution.

Let us assume that, the points X (a, -2), Y (3, 2) and Z (5, 6) are collinear.

Now,

Slope of XY = Slope of YZ

2(2)3a=6253$\Rightarrow \frac{2 – \left ( -2 \right )}{ 3 – a } = \frac{ 6 – 2 }{ 5 – 3 }$ 2+23a=42$\Rightarrow \frac{2 + 2 }{ 3 – a } = \frac{ 4 }{ 2 }$ 43a=2$\Rightarrow \frac{ 4 }{ 3 – a } = 2$

4 = 6 – 2a

2a = 2

a = 1

Therefore, the value of a = 1.

Q-9. Prove that the points (-3, -2), (5, 0), (4, 4) and (-4, 2) are the vertices of a parallelogram without using the distance formula.

Solution.

Let us denote the given points as P (-3, -2), Q (5, 0), R (4, 4) and S (-4, 2).

Now,

Slope of the side PQ = 0+25+3=28=14$\frac{ 0 + 2 }{ 5 + 3 } = \frac{ 2 }{ 8 } = \frac{ 1 }{ 4 }$

Slope of the side RS = 2444=28=14$\frac{ 2 – 4 }{ -4 – 4 } = \frac{ -2 }{ -8 } = \frac{ 1 }{ 4 }$

Thus,

Slope of the side PQ = Slope of the side RS

Hence,

PQ and RS are parallel to each other.

Now,

Slope of the side QR = 4045=41=4$\frac{ 4 – 0 }{ 4 – 5 } = \frac{ 4 }{ -1 } = -4$

Slope of the side PS = 2+24+3=41=4$\frac{ 2 + 2 }{ -4 + 3 } = \frac{ 4 }{ -1 } = -4$

Thus,

Slope of the side QR = Slope of the side PS

Hence,

QR and PS are parallel to each other.

Hence, both of the opposite side‘s pairs of the quadrilateral PQRS are parallel. Thus, PQRS is a parallelogram.

Therefore, the given points P (-3, -2), Q (5, 0), R (4, 4) and S (-4, 2) are the vertices of the parallelogram.

Q-10. Consider two points (4, -2) and (5, -3). What is the angle between the x- axis and the line joining the given two points?

Solution.

The slope of the line joining the given two points, namely, X (4, -2) and Y (5, -3) is given by:

M = 3(2)54$\frac{ -3 – \left( -2 \right ) }{ 5 – 4 }$ = -3 + 2 = -1

Thus,

The angle of inclination, say Θ$\Theta$, of the line joining the given points X (4, -2) and Y (5, -3) is:

tanΘ$\Theta$ = m = -1

Since, Θ$\Theta$ is (9045$90^{\circ} – 45^{\circ}$ ) = 135$135^{\circ}$

Therefore, between the x- axis and the line joining the given two points, the angle of inclination is 135$135^{\circ}$.

Q-11. Assume that the slope of a line AB is twice the slope of the other line PQ. Get all the slopes of the lines when the tangent of the angle between them is 13$\frac{ 1 }{ 3 }$.

Solution.

Let us assume that, the slope of the given two lines be m1 and m2, so that

m2 = 2 m1

Now,

It is known that if Θ$\Theta$ is the angle between the two given lines, namely AB and PQ, whose slopes are m1 and m2, then

tanΘ$\Theta$ = m2m11+m1m2$\left | \frac{ m_{ 2 } – m_{ 1 } }{ 1 + m_{ 1 } m_{ 2 }} \right |$

According to the question, the tangent of the angle between the two given lines is 13$\frac{ 1 }{ 3 }$.

13$\frac{ 1 }{ 3 }$ = 2m1m11+m1×2m1$\left | \frac{ 2m_{ 1 } – m_{ 1 } }{ 1 + m_{ 1 } \times 2m_{ 1 }} \right |$

13=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \left | \frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }} \right |$ 13=m11+(2m1)2or13=(m11+(2m1)2)=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }} \; or \; \frac{ 1 }{ 3 } = -\left (\frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }} \right ) = \frac{ m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }}$

CASE- I:

13=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \frac{ -m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }}$ 1+(2m1)2=3m1$\Rightarrow 1 + \left ( 2m_{ 1 } \right )^{ 2 } = -3 m_{ 1 }$ 1+3m1+(2m1)2=0$\Rightarrow 1 + 3 m_{ 1 } + \left ( 2m_{ 1 } \right )^{ 2 } = 0$ (2m1)2+2m1+m1+1=0$\Rightarrow \left ( 2m_{ 1 } \right )^{ 2 } + 2 m_{ 1 } + m_{ 1 } + 1 = 0$ 2m1(m1+1)+1(m1+1)=0$\Rightarrow 2m_{ 1 } \left ( m_{ 1 } + 1 \right ) + 1\left ( m_{ 1 } + 1 \right ) = 0$ (m1+1)(2m1+1)=0$\Rightarrow \left ( m_{ 1 } + 1 \right ) \left ( 2m_{ 1 } + 1 \right ) = 0$ m1=1orm=12$\Rightarrow m_{ 1 } = -1 \; or \; m = -\frac{ 1 }{ 2 }$

Let, m1 = -1, then the slopes of the lines are -1 and -2.

Let, m=12$m = -\frac{ 1 }{ 2 }$, then the slopes of the given lines, namely AB and PQ are 12$-\frac{ 1 }{ 2 }$ and -1.

CASE- II:

13=m11+(2m1)2$\Rightarrow \frac{ 1 }{ 3 } = \frac{ m_{ 1 } }{ 1 + \left (2m_{ 1 } \right )^{ 2 }}$ 1+(2m1)2=3m1$\Rightarrow 1 + \left ( 2m_{ 1 } \right )^{ 2 } = 3 m_{ 1 }$ 13m1+(2m1)2=0$\Rightarrow 1 – 3 m_{ 1 } + \left ( 2m_{ 1 } \right )^{ 2 } = 0$ (2m1)22m1m1+1=0$\Rightarrow \left ( 2m_{ 1 } \right )^{ 2 } – 2 m_{ 1 } – m_{ 1 } + 1 = 0$ 2m1(m11)1(m11)=0$\Rightarrow 2m_{ 1 } \left ( m_{ 1 } – 1 \right ) – 1\left ( m_{ 1 } – 1 \right ) = 0$ (m11)(2m11)=0$\Rightarrow \left ( m_{ 1 } – 1 \right ) \left ( 2m_{ 1 } – 1 \right ) = 0$ m1=1orm=12$\Rightarrow m_{ 1 } = 1 \; or \; m = \frac{ 1 }{ 2 }$

Let, m1 = 1, then the slopes of the lines are 1 and 2.

Let, m=12$m = \frac{ 1 }{ 2 }$, then the slopes of the given lines namely, AB and PQ are 12$\frac{ 1 }{ 2 }$ and 1.

Therefore, the slopes of the lines AB and PQ are -1 and -2 or,  12$-\frac{ 1 }{ 2 }$ and -1 or,  12$\frac{ 1 }{ 2 }$ and 1 or,  1 and 2.

Q-12. Consider a line passing through two points (a1, b1) and (j, k). Assume that the slope of the line passing through these points is m. Prove that:

k – b1 = m (j – a1).

Solution.

As we know that:

The slope of the line, say AB, passing through the two given points (a1, b1) and (j, k) is given by:

kb1ja1$\frac{ k – b_{ 1 } }{ j – a_{ 1 } }$

kb1ja1$\frac{ k – b_{ 1 } }{ j – a_{ 1 } }$ = m

kb1=m(ja1)$\Rightarrow k – b_{ 1 } = m \left (j – a_{ 1 } \right )$

Therefore, kb1=m(ja1)$\Rightarrow k – b_{ 1 } = m \left (j – a_{ 1 } \right )$.

Hence, proved.

Q-13. Assume that the three points (a, 0), (p, r) and (0, h) lies on the same line. Prove that:

pa+rh=1$\frac{ p }{ a } + \frac{ r }{ h } = 1$.

Solution.

As per the given condition in the question,

Points X( a, 0 ), Y( p, r ) and Z( 0, h ) lies on the same line, then

The slope of XY = The slope of YZ

r0pa=hr0p$\Rightarrow \frac{ r – 0 }{ p – a } = \frac{ h – r }{ 0 – p }$ rpa=hrp$\Rightarrow \frac{ r }{ p – a } = \frac{ h – r }{ – p }$

-pr = (h – r) (p – a)

-pr = hp – ha – rp + ra

hp + ra = ha

Dividing both side by ha, we will get

hpha+raha=haha$\frac{ hp }{ ha } + \frac{ ra }{ ha } = \frac{ ha }{ ha }$ pa+rh=1$\Rightarrow \frac{ p }{ a } + \frac{ r }{ h } = 1$

Hence, proved.

Q-14. Take the records of population and year graph given below. What will be the slope of the line XY? By using this, find the population in the year 2005.

Solution.

Here we can observe that the line XY passes through points X (1985, 90) and Y (1995, 95).

Then, the slope of the line XY is given by

959019951985=510=12$\frac{ 95 – 90 }{ 1995 – 1985 } = \frac{ 5 }{ 10 } = \frac{ 1 }{ 2 }\\$

Let, a be the population of the year 2005.

Then, as per the data in the graph, the line XY will exactly pass through the point Z( 2005, y ).

Slope of XY = Slope of YZ

12=y9520051995$\Rightarrow \frac{ 1 }{ 2 } = \frac{ y – 95 }{ 2005 – 1995 }$ 12=y9510$\Rightarrow \frac{ 1 }{ 2 } = \frac{ y – 95 }{ 10 }$ 102=y95$\Rightarrow \frac{ 10 }{ 2 } = y – 95\\$

So, y – 95 = 5

So, y = 100

Hence, the slope of the line XY is 12$\frac{ 1 }{ 2 }$, whereas, in the year 2005, the expected population will be about 100 crores.