RBSE Solutions for Class 11 Maths Chapter 11 Straight Line Ex 11.1

Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Ex 11.1

Question 1.
Find the equation of straight line which is parallel to x-axis and
(i) lie at a distance of 5 unit from origin (above origin)
(ii) lie at a distance of 3 unit from origin (below origin)
Solution:
(i) Equation of line AB

(ii) Equation of line PQ

Question 2.
Find the equations of those straight lines which are parallel to x-axis and iie at a distance :
(i) a + b
(ii) a² – b²
(iii) b cos θ
Solution:
(i) Equation of line AB

(ii) Equation of line PQ

(iii) Equation of line RS

Question 3.
Find the equation of those straight lines parallel to y-axis which are at a distance of:
(i) 5 units
(ii) -3 units
(iii) \(\frac { 2 }{ 5 } \) unit from the origin
Solution:
(i) Equation of line AB
x = 5

(ii) Equation of line PQ

(iii) Equation of line PQ

Question 4.
Find the equation of those straight line which are parallel to .y-axis and at a distance of:
(i) \(\sqrt { 7 }\)
(ii) – \(\sqrt { 3 }\)+ 2
(iii) P + q
Solution:
(i) Equation of line AB

(ii) Equation of line MN

(iii) Equation of line RS

Question 5.
Find the equations of straight lines which passes through (- 3, 2) and is perpendicular to x-axis and is parallel to x-axis respectively.
Solution:
When line is perpendicular to x-axis then will be parallel toy-axis then its equation passing through (- 3,2).
x = – 3
⇒ x + 3 = 0 (line AB)
Similarly when line is || to x-axis then equation of line passing through (- 3, 2).

Question 6.
Find the equations of lines passing through point (3,4) and parallel to both axis. Also find the equation of line parallel to these lines at a distance of 8 unit.
Solution:
Lines passing through (3, 4).
(i) Equation of line AB || to x-axis.

(ii) Equation of line PQ || toy-axis.

Let lines mn and m’n’ are situated at a distance of 8 units from AB. Then equation of line mn
y = 12 and y = – 4 or y + 4 = 0
Let lines rs and r’s’ are located at a distance of 8 unit from line PQ.

Then equations of line rs and r’s’ are
x = – 11
and x = – 5
or x + 5 = 0

Question 7.
Write the coordinates of intersection points of x = ± 4 and y = ± 3 and find the area of rectangle so formed.
Solution:
Given, x = ± 4
and y = ± 3

Equation of line AB is y = +3
Equation of line AD is x = + 4
Then coordinates of point A of intersection of lines AB and AD = (4, 3)
For the coordinates of point B
Equation of line AB ⇒ y = + 3
Equation of line BC ⇒ x = – 4
Coordinates of point B of their intersection = (- 4, 3)
For the coordinates of point C,
Equation of line BC ⇒ x = – 4
Equation of line CD ⇒ y = – 3
Coordinates of point C of their intersection = (- 4, – 3)
For the coordinates of point D Equation of line CD ⇒ y = – 3
Equation of line AD ⇒ x = + 4
Coordinates of point D of their intersection = (4, – 3)
Thus coordinates of points, B, C and D are respectively. (4, 3), (- 4, 3), (- 4, – 3) and (4,- 3)
Then,

Question 8.
Find the equations of those lines which passes through origin and
(i) makes an angle of – 135° with x-axis.
(ii) makes an angle of 60° with OY in Ist quadrant.
(iii) cut intercepts of 5 units with +ve axis of y and is parallel to bisector of angle XOY.
Solution:
Standard equation of line with slope m and passes through origin,
y = mx

Question 9.
Find the equations of those lines which cuts the following intercepts at x-axis and .y-axis.
(i) 5, 3
(ii) – 2, 3
Solution :
(i) Equation of line PQ

(ii) Equation of line PQ

Question 10.
Find the equation of the line which passes through (2, 3) and cuts equal intercepts on both axis.
Solution:
Equation of line AB

\(\frac { x }{ a } \) + \(\frac { y }{ a } \) = 1
Since this line passes through the point (2, 3).
∴ 2 + 3 = a
⇒ a = 5
Thus x + y = 5

Question 11.
Find the equation of straight line which passes through point (1,2) and cut intercepts on x-axis which is twice the intercepts on y-axis.
Solution :
Straight line passes through point (1,2) and intercept at x-axis is doubled the intercept by y-axis.
Equation of line AB

Question 12.
Find the equation of the lines which passes through point (- 3, – 5) and intercepts cut by line between both axis bisect this point.
Solution:
Point (- 3, – 5) lies on line PQ which bisects it. Thus intercepts of x and y-axis will be – 6 and – 10.

⇒ 10x + 6y + 60 = 0
⇒ 5x + 3y + 30 = 0

Question 13.
Find the equation of two lines which passes through point (4, – 3) and sum of intercepts cut by axis is 5 unit.
Solution:
Let equation of straight line in intercept form is as follows :
\(\frac { x }{ a } \) + \(\frac { y }{ b } \) = 1
or \(\frac { x }{ a } \) + \(\frac { y }{ 5-a } \) = 1
This line passes through point (4, – 3).
⇒ \(\frac { 4 }{ a } \) – \(\frac { 3 }{ 5-a } \) = 1
⇒ 20 – 4a – 3a = 5a – a²
⇒ 20 – 7a – 5a + a² = 0
⇒ a² – 12a + 20 = 0
⇒ a² – 10a – 2a + 20 = 0
⇒ a(a – 10) -2(a – 10) = 0
⇒ (a – 2) (a – 10) = 0
When a – 2 = 0
Then a = 2

Question 14.
Prove that equation of line at which axis reciprocals of intercepts are a and b is ax + by = 1.
Solution:
Equation of line PQ in form of intercepts,

Question 15.
A straight line cuts intercepts with axis 5 unit and 3 unit respectively. Find the equation of line where intercepts is :
(i) in +ve direction of axis
(ii) in -ve direction of axis
(iii) first intercept in +ve and second in -ve direction
Solution:
(i)

Equation of line
\(\frac { x }{ 5 } \) + \(\frac { y }{ 3 } \) = 1
3x + 5y – 15 = 0

Equation of line
\(\frac { x }{ -5 } \) + \(\frac { y }{ -3 } \) = 1
⇒ – 3x – 5y – 15 = 0
⇒ 3x + 5y + 15 = 0

Equation of line
\(\frac { x }{ 5 } \) + \(\frac { y }{ -3 } \) = 1
⇒ – 3x + 5y + 15 = 0
⇒ 3x – 5y – 15 = 0

Question 16.
The perpendicular drawn from origin to a straight line makes an angle of 30° with .y-axis and its length is 2 units. Find the equation of this line. Solution:
According to question,
p = 2 units
0 = 180° – 30°= 150°

Question 17.
Find the length of that part of line x sin a + y cos a = sin 2α which cuts axis at mid point.Also, find the coordinate of mid point of this part.
Solution :
x sin α + y cos α = sin 2α
⇒ x sin α + y cos α = 2 sin α cos α

Thus intercept of x – axis = 2 cos α and intercept aty-axis = 2 sin α and coordinates of point A and B are (2 cos α, 0) and (0, 2 sin α)

Thus, length of middle part cut of axis by given line is 2 unit and coordinates of mid point are (cos α, sin α).

Question 18.
Find the equation of straight line for which p = 3 and cos α = \(\frac { \sqrt { 3 } }{ 2 } \), where p is the length of perpendicular drawn from origin to the line and α is the angle formed by perpendicular with x-axis.
Solution:
Given
P= 3

RBSE Solutions for Class 11 Maths