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# Straight Lines - NCERT Solution, Class 11 Mathematics Class 11 Notes | EduRev

## Class 11 : Straight Lines - NCERT Solution, Class 11 Mathematics Class 11 Notes | EduRev

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Class XI Chapter 10 – Straight Lines Maths

Page 1 of 68
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.
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,
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 (x
1
, y
1
), (x
2
, y
2
), and (x
3
, y
3
) is

Therefore, area of ?ABC
Page 2

Class XI Chapter 10 – Straight Lines Maths

Page 1 of 68
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.
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,
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 (x
1
, y
1
), (x
2
, y
2
), and (x
3
, y
3
) is

Therefore, area of ?ABC

Class XI Chapter 10 – Straight Lines Maths

Page 2 of 68

Area of ?ACD

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 3

Class XI Chapter 10 – Straight Lines Maths

Page 1 of 68
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.
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,
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 (x
1
, y
1
), (x
2
, y
2
), and (x
3
, y
3
) is

Therefore, area of ?ABC

Class XI Chapter 10 – Straight Lines Maths

Page 2 of 68

Area of ?ACD

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).

Class XI Chapter 10 – Straight Lines Maths

Page 3 of 68

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
+ a
2

? 4a
2
– a
2
= (OA)
2

? (OA)
2
= 3a
2

? OA =
?Coordinates of point A =
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and or
(0, a), (0, –a), and .

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 are and .
(i) When PQ is parallel to the y-axis, x
1
= x
2
.
Page 4

Class XI Chapter 10 – Straight Lines Maths

Page 1 of 68
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.
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,
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 (x
1
, y
1
), (x
2
, y
2
), and (x
3
, y
3
) is

Therefore, area of ?ABC

Class XI Chapter 10 – Straight Lines Maths

Page 2 of 68

Area of ?ACD

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).

Class XI Chapter 10 – Straight Lines Maths

Page 3 of 68

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
+ a
2

? 4a
2
– a
2
= (OA)
2

? (OA)
2
= 3a
2

? OA =
?Coordinates of point A =
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and or
(0, a), (0, –a), and .

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 are and .
(i) When PQ is parallel to the y-axis, x
1
= x
2
.

Class XI Chapter 10 – Straight Lines Maths

Page 4 of 68
In this case, distance between P and Q

(ii) When PQ is parallel to the x-axis, y
1
= y
2
.
In this case, distance between P and Q

Question 4:
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Answer
Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

On squaring both sides, we obtain
a
2
– 14a + 85 = a
2
– 6a + 25
? –14a + 6a = 25 – 85
? –8a = –60

Thus, the required point on the x-axis is .

Question 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).
Answer
Page 5

Class XI Chapter 10 – Straight Lines Maths

Page 1 of 68
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.
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,
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 (x
1
, y
1
), (x
2
, y
2
), and (x
3
, y
3
) is

Therefore, area of ?ABC

Class XI Chapter 10 – Straight Lines Maths

Page 2 of 68

Area of ?ACD

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).

Class XI Chapter 10 – Straight Lines Maths

Page 3 of 68

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
+ a
2

? 4a
2
– a
2
= (OA)
2

? (OA)
2
= 3a
2

? OA =
?Coordinates of point A =
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and or
(0, a), (0, –a), and .

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 are and .
(i) When PQ is parallel to the y-axis, x
1
= x
2
.

Class XI Chapter 10 – Straight Lines Maths

Page 4 of 68
In this case, distance between P and Q

(ii) When PQ is parallel to the x-axis, y
1
= y
2
.
In this case, distance between P and Q

Question 4:
Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Answer
Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

On squaring both sides, we obtain
a
2
– 14a + 85 = a
2
– 6a + 25
? –14a + 6a = 25 – 85
? –8a = –60

Thus, the required point on the x-axis is .

Question 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).
Answer

Class XI Chapter 10 – Straight Lines Maths

Page 5 of 68

The coordinates of the mid-point of the line segment joining the points
P (0, –4) and B (8, 0) are
It is known that the slope (m) of a non-vertical line passing through the points (x
1
, y
1
)
and (x
2
, y
2
) is given by .
Therefore, the slope of the line passing through (0, 0) and (4, –2) is
.
Hence, the required slope of the line is .

Question 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.
Answer
The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).
It is known that the slope (m) of a non-vertical line passing through the points (x
1
, y
1
)
and (x
2
, y
2
) is given by .
?Slope of AB (m
1
)
Slope of BC (m
2
)
Slope of CA (m
3
)
It is observed that m
1
m
3
= –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.

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