NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines Ex 10.1
Page No 211:
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
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_m12906fa2.jpg)
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (ΔABC) + area (ΔACD)
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_2cd45cb1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_2cd45cb1.gif)
Therefore, area of ΔABC
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_m7d03e54d.gif)
Area of ΔACD
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_m45f3e384.gif)
Thus, area (ABCD)![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_m12dba1d2.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4449/chapter%2010_html_m12dba1d2.gif)
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).
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.
![](https://img-nm.mnimgs.com/img/study_content/editlive_ncert/58/2012_11_05_12_36_14/1.png)
On applying Pythagoras theorem to ΔAOC, we obtain
(AC)2 = (OA)2 + (OC)2
⇒ (2a)2 = (OA)2 + a2
⇒ 4a2 – a2 = (OA)2
⇒ (OA)2 = 3a2
⇒ OA =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4462/chapter%2010_html_ma31b01e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4462/chapter%2010_html_ma31b01e.gif)
∴Coordinates of point A =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4462/chapter%2010_html_dbebbe9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4462/chapter%2010_html_dbebbe9.gif)
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and
or (0, a), (0, –a), and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4462/chapter%2010_html_3b3eecc1.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4462/chapter%2010_html_m6df5e672.gif)
Question 3:
Find the distance between
and
when: (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m2ddc22e5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m7c277b47.gif)
Answer:
The given points are
and
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m2ddc22e5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m7c277b47.gif)
(i) When PQ is parallel to the y-axis, x1 = x2.
In this case, distance between P and Q ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m13876f44.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m13876f44.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_5c17477.gif)
(ii) When PQ is parallel to the x-axis, y1 = y2.
In this case, distance between P and Q ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m13876f44.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_m13876f44.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4466/chapter%2010_html_me9260d9.gif)
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).
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4472/chapter%2010_html_m6fbcd0af.gif)
On squaring both sides, we obtain
a2 – 14a + 85 = a2 – 6a + 25
⇒ –14a + 6a = 25 – 85
⇒ –8a = –60
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4472/chapter%2010_html_m1ad19f4f.gif)
Thus, the required point on the x-axis is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4472/chapter%2010_html_m16a471b5.gif)
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:
The coordinates of the mid-point of the line segment joining the points
P (0, –4) and B (8, 0) are![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4475/chapter%2010_html_587f9983.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4475/chapter%2010_html_587f9983.gif)
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4475/chapter%2010_html_517b2d17.gif)
Therefore, the slope of the line passing through (0, 0) and (4, –2) is
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4475/chapter%2010_html_m2edf2092.gif)
Hence, the required slope of the line is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4475/chapter%2010_html_5c647a5.gif)
Page No 212:
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 (x1, y1) and (x2, y2) is given by
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_517b2d17.gif)
∴Slope of AB (m1)![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_42f854ed.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_42f854ed.gif)
Slope of BC (m2)![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_7bc29b13.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_7bc29b13.gif)
Slope of CA (m3)![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_m434e3f70.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4480/chapter%2010_html_m434e3f70.gif)
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.
Question 7:
Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Answer:
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°.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4489/chapter%2010_html_37f2d159.jpg)
Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60° ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4489/chapter%2010_html_1fe283ac.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4489/chapter%2010_html_1fe283ac.gif)
Question 8:
Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.
Answer:
If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then
Slope of AB = Slope of BC
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4494/chapter%2010_html_57733e18.gif)
Thus, the required value of x is 1.
Question 9:
Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and
(–3, 2) are vertices of a parallelogram.
Answer:
Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_6165529d.jpg)
Slope of AB![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_1d2e945.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_1d2e945.gif)
Slope of CD =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_m28501ce6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_m28501ce6.gif)
⇒ Slope of AB = Slope of CD
⇒ AB and CD are parallel to each other.
Now, slope of BC =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_14869458.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_14869458.gif)
Slope of AD =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_53a10719.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4500/chapter%2010_html_53a10719.gif)
⇒ 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.
Question 10:
Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).
Answer:
The slope of the line joining the points (3, –1) and (4, –2) is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4509/chapter%2010_html_1f8e1a33.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4509/chapter%2010_html_1f8e1a33.gif)
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°.
Question 11:
The slope of a line is double of the slope of another line. If tangent of the angle between them is
, find the slopes of he lines.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_18a817bc.gif)
Answer:
Let
be the slopes of the two given lines such that
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_m4958eda3.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_5b021876.gif)
We know that if θisthe angle between the lines l1 and l2 with slopes m1 and m2, then
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_56e8bd97.gif)
It is given that the tangent of the angle between the two lines is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_18a817bc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_m64ebd05b.gif)
Case I
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_4e50b3cf.gif)
If m = –1, then the slopes of the lines are –1 and –2.
If m =
, then the slopes of the lines are
and –1.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_m7dfc0570.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_m7dfc0570.gif)
Case II
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_5c6014f6.gif)
If m = 1, then the slopes of the lines are 1 and 2.
If m =
, then the slopes of the lines are
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_eeecab0.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_757e9d3d.gif)
Hence, the slopes of the lines are –1 and –2 or
and –1 or 1 and 2 or
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_m7dfc0570.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4512/chapter%2010_html_757e9d3d.gif)
Question 12:
A line passes through
. If slope of the line is m, show that
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_me35eed5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_m7c9e09b6.gif)
Answer:
The slope of the line passing through
is
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_me35eed5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_m1a3bb387.gif)
It is given that the slope of the line is m.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_1fadf1e0.gif)
Hence,![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_m7c9e09b6.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4516/chapter%2010_html_m7c9e09b6.gif)
Question 13:
If three point (h, 0), (a, b) and (0, k) lie on a line, show that
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4525/chapter%2010_html_3543ed7c.gif)
Answer:
If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then
Slope of AB = Slope of BC
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4525/chapter%2010_html_m7c9d9416.gif)
On dividing both sides by kh, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4525/chapter%2010_html_m1e6c0152.gif)
Hence,![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4525/chapter%2010_html_3543ed7c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4525/chapter%2010_html_3543ed7c.gif)
Question 14:
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?
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4557/chapter%2010_html_m5ddbf1ab.jpg)
Answer:
Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4557/chapter%2010_html_71b86fa5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4557/chapter%2010_html_71b86fa5.gif)
Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y).
∴Slope of AB = Slope of BC
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4557/chapter%2010_html_a8ea215.gif)
Thus, the slope of line AB is
, while in the year 2010, the population will be 104.5 crores.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/170/4557/chapter%2010_html_eeecab0.gif)