# 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 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. 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°. 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. 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 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? 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.