## 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) a^{2} – b^{2}

(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 – a^{2}

⇒ 20 – 7a – 5a + a^{2} = 0

⇒ a^{2} – 12a + 20 = 0

⇒ a^{2} – 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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