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
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