Straight Lines – Exercise 10.1 – Class XI

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

Solution:

Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).

Now let us plot A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as

Straight lines - Exercise 10.1 - Class XI

To find the area of quadrilateral ABCD, we draw one diagonal, say AC.

Thus, area (ABCD) = area (∆ABC) + area (∆ACD)

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is

1/2|x1(y2 – y3)+ x2(y3 – y1)+ x3(y1 – y2)|

Therefore, area of ∆ABC

= 1/2|-4(7+5)+0(-5-5)+5(5-7)|unit2

= 1/2|-4(12)+5(-2)| unit2

= 1/2|-48-10|unit2

= 1/2|-58|unit2

= 1/2 x 58 unit2

= 29 unit2

Area of ∆ACD

= 1/2|-4(-5 + 2)+5(-2 – 5)+(-4)(5+5)|unit2

= 1/2|-4(-3)+5(-7)-4(10)| unit2

= 1/2|12 – 35 – 10|unit2

= 1/2|-63|unit2

= 1/2 x 63 unit2

= 63/2 unit2

Thus, area(ABCD) = (29 + 63/2) unit2 = (58+63)/2 unit2 = 121/2 unit2


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

Solution:

Let ABC be the given equilateral triangle with side 2a.

Accordingly, AB = BC = CA = 2a

Let us assume base BC lies along the y-axis such that the mid-point of BC is at the origin.

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on the y-axis.

Straight lines - Exercise 10.1 - Class XI

On applying Pythagoras theorem to ∆AOC, we obtain

(AC)2 = (OA)2 + (OC)2

⇒ (2a)2 = (OA)2 + a2

⇒ 4a2 – a2 = (OA)2

⇒ (OA)2 = 3a2

⇒ OA =√3a

∴Coordinates of point A = (±√3a, 0)

Therefore, the vertices of the given equilateral triangle are

(0, a), (0, –a), and (√3a, 0)

or

(0, a), (0, –a), and (-√3a, 0)


  1. 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.

Solution:

The given points are P(x1, y1) and Q(x2, y2)

(i) When PQ is parallel to the y-axis, x1 = x2.

In this case, distance between P and Q =  √[(x2 – x1)2+(y2 – y1)2]

= √(y2 – y1)2

=|y2 – y1|

(ii)When PQ is parallel to the x-axis, y1 = y2.

In this case, distance between P and Q=  √[(x2 – x1)2+(y2 – y1)2]

= √(x2 – x1)2

=|x2 – x1|


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

Solution:

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

√[(7-a)2+(6-0)2] = √[(3-a)2 +(4 – 0)2]

⇒ √[49+a2 -14a +36] = √[9+a2 – 6a + 16]

√[a2 – 14a + 85] = √[a2 – 6a + 25]

On squaring both sides, we obtain a2 – 14a + 85 = a2 – 6a + 25

⇒ –14a + 6a = 25 – 85

⇒ –8a = –60

a = 60/8 = 15/2

Thus, the required point on the x-axis is (15/2, 0).


  1. 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).

Solution:

The coordinates of the mid-point of the line segment joining the points P (0, –4) and B (8, 0) are (0+8/2 , -4+0/2) = (4, -2)

It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by m = (y2 – y1)/(x2 – x1) , x2 ≠ x1

Therefore, the slope of the line passing through (0, 0) and (4, –2) is

-2-0/4-0 = -2/4 = 1/2

Hence, the required slope of the line is –1/2.


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

Solution:

Given: The vertices of the triangle are A (4, 4), B (3, 5), and C (–1, –1).

We know that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is m = (y2 – y1)/(x2 – x1) , x2 ≠ x1

Slope of AB (m1) = 5-4/3-4 = -1

Slope of BC (m2) = -1-5/-1-3 = -6/-4 = 3/2

Slope of CA (m3) =  4+1/4+1  = 5/5 = 1

It is observed that m1m3 = –1

This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.


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

Solution:

If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° + 30° = 120°.

Straight Lines - Exercise 10.1 - Class XI

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60° = – √3


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

Solution:

If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then

Slope of AB = Slope of BC

1-(-1)/2-x = 5-1/4-2

1+1/2-x = 4/2

2/2-x = 2

2 = 4 – 2x

2x = 2

x = 1

Thus, the required value of x is 1.


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

Solution:

Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

Slope of AB = 0+1/4+2 = 1/6

Slope of AB = 2-3/-3-3 = -1/-6= 1/6

Slope of AB = Slope of CD

Thus AB and CD are parallel to each other

Now slope of BC = 3-0/3-4 = 3/-1 = -3

Slope of AD = 2+1/-3+2 = 3/-1 = – 3

⇒ Slope of BC = Slope of AD

⇒ BC and AD are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.

Straight Lines - Exercise 10.1 - Class XI

Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.


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

Solution:

The slope of the line joining the points (3, –1) and (4, –2) is m = -2-(-1)/4-3 = -2 + 1 = -1

Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by tan θ = –1

⇒ θ = (90° + 45°) = 135°

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.


  1. The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3, find the slopes of the lines.

Solution:

Let m1 and m be the slopes of the two given lines such that m1 = 2m.

We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then

Straight Lines - Exercise 10.1 - Class XI

 

1+ 2m2 = 3m

2m2 – 3m + 1 = 0

2m2–  2m – m + 1 = 0

2m(m – 1) – 1(m – 1) = 0

(m – 1)(2m – 1) = 0

m = 1 or m = 1/2

If m = 1 then the slopes of the lines are 1 and 2

IF m = 1/2 then the slopes of the lines are  1/2 and  1

Hence the slopes of the lines are -1 and -2 or –1/2 and -1 or 1 and 2 or 1/2 and 1


  1. A line passes through (x1, y1). IF slope of the line is m1, show that k – y1 = m(h – x1)

Solution:

The slope of the line passing through (x1, y1) and (h, k) is (k – y1)/ (h – x1)

It is given that the slope of the line is m

[(k – y1)/ (h – x1)] = m

k – y1 = m(h – x1)


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

Solution:

IF the points A(h, 0) , B(a, b) and C(0, k) lie on a line then

Slope of AB = slope of BC

b-0/a-h = k-h/0-a

b/a-h = k-h/-a

-ab = (k – b)(a – h)

-ab = ka – kh – ab + bh

ka + bh = kh

On dividing both sides by kh, we obtain

ka/kh + bh/kh = kh/kh

a/h + b/k = 1

Therefore, a/h + b/k = 1


  1. Consider the given population and year graph. Find the slope of the line AB and using it, find what will be the population in the year 2010?

Straight Lines - Exercise 10.1 - Class XI

Solution:

Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is

97-92/1995-1985 = 5/10= 1/2

Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y).

Therefore, Slope of AB = Slope of BC

1/2 = y-97/2010 – 1995

1/2 = y-97/15

15/2 = y – 97

y – 97 = 7.5

y = 7.5 + 97 = 104.5

Thus the slope of the line AB is 1/2 while in the year 2010, the population will be 104.5 crores.