Chapter 10 Straight Lines
ALL CBSE STUFF
Class XI
Chapter 10 – Straight Lines
Maths
Exercise 10.1
Question 1:
Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.
14
Answer
Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–
4, –2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA,
SU
M
IT
©
20
the given quadrilateral can be drawn as
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, 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
Therefore, area of ∆ABC
Page 1 of 68
Chapter 10 – Straight Lines
Maths
20
14
Class XI
IT
©
Area of ∆ACD
SU
M
Thus, area (ABCD)
Question 2:
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.
Answer
Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that 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).
Page 2 of 68
Class XI
Chapter 10 – Straight Lines
Maths
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
20
14
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ∆AOC, we obtain
⇒ (2a)2 = (OA)2 + a2
⇒ 4a2 – a2 = (OA)2
⇒ OA =
IT
⇒ (OA)2 = 3a2
©
(AC)2 = (OA)2 + (OC)2
∴Coordinates of point A =
SU
M
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and
(0, a), (0, –a), and
or
.
Question 3:
Find the distance between
and
when: (i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.
Answer
The given points
Class XI
Chapter 10 – Straight Lines
Maths
Exercise 10.1
Question 1:
Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.
14
Answer
Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–
4, –2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA,
SU
M
IT
©
20
the given quadrilateral can be drawn as
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, 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
Therefore, area of ∆ABC
Page 1 of 68
Chapter 10 – Straight Lines
Maths
20
14
Class XI
IT
©
Area of ∆ACD
SU
M
Thus, area (ABCD)
Question 2:
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.
Answer
Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that 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).
Page 2 of 68
Class XI
Chapter 10 – Straight Lines
Maths
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
20
14
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ∆AOC, we obtain
⇒ (2a)2 = (OA)2 + a2
⇒ 4a2 – a2 = (OA)2
⇒ OA =
IT
⇒ (OA)2 = 3a2
©
(AC)2 = (OA)2 + (OC)2
∴Coordinates of point A =
SU
M
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and
(0, a), (0, –a), and
or
.
Question 3:
Find the distance between
and
when: (i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.
Answer
The given points