RBSE Solutions for Class 11 Maths Chapter 11 Straight Line

RBSE Solutions for Class 11 Maths Chapter 11 Straight Line

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) 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) √7
(ii) – √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

x/a + 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 :
x/a + y/b = 1
or x/a + y/5−a = 1
This line passes through point (4, – 3).
⇒ 4/a – 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
x/5 + y/3 = 1
3x + 5y – 15 = 0

Equation of line
x/−5 + y/−3 = 1
⇒ – 3x – 5y – 15 = 0
⇒ 3x + 5y + 15 = 0

Equation of line
x/5 + 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 = √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

Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Ex 11.2

Question 1.
Convert the following equations in slope form and intercept form and find the value of constants used in standard form.
(i) 7x – 13y = 15
(ii) 5x + 6y + 8 = 0
Solution:

Question 2.
Find the slope of line x cos α + y sin α = p
Solution :
x cos α + y sin α =p
⇒ y sin α = – x cos α – p

Question 3.
Find the tangent of the following lines with +ve x-axis.
(i) √3x – y + 2 = 0
(ii) x + √3y – 2 √3 = 0
Solution:
(i) √3x – y + 2 = 0
⇒ y = √3x + 2
Comparing with y = mx + c
m = √3
⇒ m = tan θ
tan θ = √3 = tan 60°
Thus tangent of required angle is 60°.

Question 4.
Prove that (x₁, y₁) will be coordinate of mid point of section cut by line

Solution:
Let straight line cuts x and y axis of points B and A and coordinates of points A and B are (0, 2y₁)

Question 5.
Find the length of intercept cut between the axis from straight line 3x + 4y = 6 and also find its mid point.
Solution:
3x + 4y = 6
⇒ 3x/6 + 4y/6 = 1

Comparing this equation x/a + y/b = 1, a = 2 and b = 3/2
Straight line intersect the axis at the points (0,3/2) at and (2,0)

Question 6.
Find the values of a and b where equations 5x – 4y = 20 and ax – by + 1 = 0 represent same straight line.
Solution:

Question 7.
Reduce the following equations in the form x cos α + y sin α = p.
(i) x + y + √2 = 0
(ii) √3x – y + 2 = 0
Solution:
(i) x + y + √2 = 0
⇒ x + y = –√2
⇒ – x – y = √2
Dividing both side by


Thus perpendicular form of given equation.
x cos 225° + 7 sin 225° = 1

Thus perpendicular form of given equation
x cos 150° + y sin 150° = 1

Question 8.
Reduce the straight line 3x – 4y – 11 = 0 in the normal form and find length of perpendicular from origin to line and its slope from x-axis.
Solution:
3x – 4y – 11 = 0
3x – 4y = 11

Question 9.
x/a + y/b = 1 and 2x – 3y = 5 represent same line, then find the values of a and b.
Solution :

Question 10.
If straight line y = mx + c and x cos α + 7 sin α = p represent same line, then find slope of line with x axis, and length of intercept cut from y – axis.
Solution:
y = mx+c …(i)
x cos α + 7 sin α = p
⇒ 7 sin α = -x cos α + p

Comparing equation (i) and (ii)
m = tan θ = -cot α
⇒ m = tan θ = tan (90° + α)
Thus θ = 90° + α
Slope of line = 90° + α


Thus slope of line is 90° + α and intercept is p cosec α.

Question 11.
Find the equation of straight line which passes through point (2, 3) and makes an angle of 45° with x- axis.
Solution :
Here point (x₁,y₁) = (2,3) and slope θ = 45°
⇒ tan θ = tan 45° = 1
∴ Equation of line y – y₁ = m (x – x₁)
⇒ y – 3 = 1 (x – 2)
⇒ y – 3 = x – 2
⇒ x – y + 1 = 0

Question 12.
Find the equation of line passing through the following two given points :
(i) (3, 4) and (5, 6)
(ii) (0, – a) and (b, 0)
(iii) (a, b) and (a + b, a- b)
(iv) (at₁, alt₁) and (at₂, alt₂)
(v) (a sec α, b tan α) and (α sec β, b tan β)
Solution:
(i) Equation of line passing through (3, 4) and (5, 6)

(ii) Equation of line passing through (0, – a) and (b, 0)

(iii) Equation of line passing through (a, b) and (a + b, a – b)

(iv) Equation of line passing through (at₁ alt₁) and (at₂, alt₂)

(v) Equation of line passing through (a sec α, b tan α) and (a sec β, b tan β)

Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Ex 11.3

Question 1.



From equation (i)
m₁ = tan θ₁ = b/a
From equation (ii)
m₂ = tan θ₂ = a/b
Therefore, angle between both lines is

Question 2.
Prove that following straight lines are parallel
(i) 2y = mx + c and 4y = 2mx
(ii) x cos α + y sin α = p and x + y tan α = 5 tan α
Solution:
(i) 2y = mx + c

Comparing with y = m₁x + c₁
m₁ = m/2
4y = 2 mx

Thus, both lines will be parallel.

(ii) x cos α + y sin α = p
⇒ y sin α = – x cos α + p

We see that
m₁ = m₂ = – cot α
Therefore both line will be parallel.

Question 3.
Prove that lines whose equation are 4x + 5y + 7 = 0 and 5x – 4y – 11 = 0, are perpendicular to each other.
Solution:
4x + 5y + 7 = 0
⇒ 5y = -4x – 7

Therefore both lines will be perpendicular.

Question 4.
Find the equations of those straight lines which
(i) Passing through point (4,5) and is parallel to line 2x – 3y – 5 = 0.
(ii) Passing through point (1,2) and is perpendicular to line 4x + 3y + 8 = 0.
(iii) Is parallel to line 2x + 5y = 7 and passes through the mid point of line joining the points (2,7) and (-4,1).
(iv) Divide the line joining the points (- 3, 7) and (5, – 4) in the ratio 4 : 7 and is perpendicular to this.
Solution:
(i) Equation of line parallel to line
2x – 3y – 5 = 0
2x – 2y + c = 0
It passes through point (4,5)
then 2 × 4 -3 × 5 + c = 0
⇒ 8 – 15 + c = 0
⇒ c = 7
Hence, equation of straight line is
2x – 3y + 7 = 0
(ii) Equation of line perpendicular to line
4x + 3y + 8 = 0
3x – 4y + c = 0
This line passes through point (1,2), then
3 × 1 – 4 × 2 + c = 0
⇒ 3 – 8 + c = 0
⇒ c = 5
Thus, equation to straight line is
3x – 4y + 5 = 0
(iii) Equation of line parallel to line
2x + 5y = 7
2x + 5y = c …(i)
Mid point of line joining the points (2,7) and (-4, 1)

Putting this value in equation (i),
⇒ 2 × (-1) + 5 × (4) = c
⇒ -2 + 20 = c
⇒ c = 18
Thus, Required equation of line is
2x + 5y = 18
(iv) Equation of line passing through the points (- 3, 7) and (5, – 4)

Equation of line perpendicular to this line
11y – 8x =c …(i)
Coordinates of point which divide the line joining the points (-3, 7) and (5, -4) in the ratio 4 : 7

Line passes through this point so putting this point in equation (i)

Thus, equation of required line
11y – 8x = 371/11
⇒ 121y – 88x = 371
⇒ 88x – 121y + 371 = 0

Question 5.
The vertices of a triangle are (0, 0), (4, – 6) and (1, – 3) Find the equations of perpendiculars drawn from these points to the corresponding sides.
Solution:
Let altitudes of ∆ABC are AD, BE, CF then AD ⊥BC, BE ⊥ CA and CF ⊥ AB Now slope of side BC

Equation of perpendicular AD Which passes through point (0, 0)
y – 0 = 1(x – 0)
⇒ y = x
⇒ y – x = 0
Equation of perpendicular BE, which passes through point (4, – 6)
y + 6= 1/3 (x – 4)
⇒ 3y + 18 – x + 4 = 0
⇒ 3y – x + 22 = 0
⇒ x – 3y = 22
Equation of perpendicular CF which passes through point (1,-3)
y + 3 = 2/3 (x – 1)
⇒ 3y + 9 – 2x + 2 = 0
⇒ 2x – 3y = 11

Question 6.
Find the ortho center of triangle whose vertices are (2, 0), (3, 4) and (0, 3).
Solution:
Let vertices of triangle are A(2, 0), B(3, 4) and C(0, 3)

Multiply equation (i) by 4 and subtracting this from equation (ii).

Question 7.
Two vertices of a triangle are (3, -1) and (- 2. 3). Orthocentre of triangle lies at the origin. Find the coordinates of third vertex.
Solution:
Vertices of ∆ABC are A(h, k), 5(3, – 1) and (- 2,3). vertex A is intersecting point of AB and AC. Side AB passes through point (3, – 1) and is perpendicular to DC whose slope is – 3/2

Similarly AC passes through (- 2,3) and is perpendicular to OB, whose slope is – 1/3. Thus equation of side AC
y – 3 = 3(x + 2)
⇒ 3x – y + 9 = 0 …(ii)
Solving equation (i) and (ii)
x = – 36/7 and y = – 45/7
Thus, coordinates of A is ( – 36/7 – 45/7)

Question 8.
Find the equation of perpendicular bisector qf line joining the points (2, -3) and (- 1, 5).
Solution:
AB is a line joining the points A(2, -3) and B(-1, 5)

Equation of line perpendicular to this line

Question 9.
Find the equation of line which is perpendicular on straight line x/a – y/b = 1 at the point where it meets the x-axis
Solution :
Let equation of line AB is x/a – y/b =1, then co-ordinates of that point where it meets with x-axis will be (a, 0). We get equation of line ⊥ to this line.

Question 10.
Find the equation of the line which is parallel to the line 2x + 3y + 11 = 0 and sum of intercepts cut by axis is 15.
Solution:
Equation of given line
2x + 3y + 11 = 0
Equation of line parallel to this line.
2x + 3y + c = 0 ….(i)
⇒ 2x + 3y = -c

Question 11.
Find the equation of straight lines which passes through the point (2, – 3) and makes an angle of 45° with line 3x – 2y = 4.
Solution:
Equation of line passing through point (2, -3)
y + 3 = m₁ (x – 2)
⇒ y = m₁ (x – 2) -3
⇒ y = m₁ x – (2m₁ + 3)
Eqⁿ of given straight line
3x – 2y = 4
⇒ 2y = 3x – 4


Thus, equation of line
y + 3 = – 5 (x – 2)
⇒ y + 3 + 5(x – 2) = 0
⇒ y + 5x – 7 = 0

Thus, equation of line
y + 3 = 1/5 (x – 2)
⇒ 5y + 15 – x + 2 = 0
⇒ 5y – x + 17 = 0
Thus, equation of lines are
y + 5x – 7 = 0 and 5y – x + 17 = 0

Question 12.
Find the equation of straight line which passes through the point (4, 5) and make similar angles with lines 3x = 4y + 7 and 5y = 12x + 6
Solution:
Equation of line passing through the point (4, 5)
y – 5 = m₁ (x – 4) …(i)
Given line 3x = 4y + 7
⇒ 4y = 3x – 7
⇒ y = 3/4 x – 7/4 …(ii)
Comparing with y = m₂ x + c₂
m₂ = 3/4
other line 5y = 12x + 6
⇒ y = 12/5 x + 6/5 …(iii)
Comparing with y = m₃x + c₃
According to question m₃ = 12/5
Angle between line (i) and (ii) = Angle between line (i) and (iii)

Question 13.
Prove that following will be equation of line which passes through origin and makes angle θ with line y = mx + c

Solution:
Let equation of line PQ is y = mx + c and it makes angle θ with PQ and its gradient = m₁

Question 14.
Prove that equation of line which passes through a point (a cos³ θ, a sin³ θ) and is perpendicular to the x sec θ + y cosec θ = a is x cos θ – y sin θ = a cos 2θ
Solution:
Line x sec θ + y cosec θ = a

⇒ y sin θ – a sin⁴ θ = x cos θ – a cos⁴ θ
⇒ x cos θ – y sin θ = a cos⁴ θ – a sin⁴ θ
⇒ x cos θ – y sin θ = a (cos⁴ θ – sin⁴ θ)
⇒ xcos θ – y sin θ = a (cos² θ + sin² θ)
(cos² θ – sin² θ)
⇒ x cos θ – y sin θ = a × 1 × cos2θ
⇒ x cos θ – y sin θ = a cos2θ
Thus equation of required perpendicular line
x cos θ – y sin θ = a cos2θ

Question 15.
Vertex of an equilateral triangle is (2, 3) and equation of its opposite side is x + y = 2. Find the equation of the remaining sides.
Solution:
Let, In ∆ABC, coordinate of A are (2,3) and equation of BC is x + y = 2. Equation of line passes through point A (2,3) and making angle of 60° with x + y = 2 is.

Question 16.
Find the equation oftwo lines which pass through point (3,2) and makes an angle of 60° with line x + √3y = 1.
Solution:
We know that equation of a line passing through point (h, k) and makes angle α with line y = mx + c is

Rajasthan Board RBSE Class 11 Maths Chapter 11 Straight Line Miscellaneous Exercise

Question 1.
AB is a line joining the points A(2, -3) and B(-1, 5) Equation of line perpendicular to this line
(A) y = 5
(B) x = 5
(C) x = -5
(D) y = -5
Solution:
Equation of line parallel to y-axis is x = -5

Thus, option (c) is correct.

Question 2.
Equation of a line which passes through point (3, – 4) and is parallel to x-axis is :
(A) x – 3
(B) y = -4
(C) x + 3 = 0
(D) y – 4 = 0
Solution:
Equation of line parallel to x-axis is y = -4

Thus, option (b) is correct.

Question 3.
Slope of y-axis is :
(A) 1
(B) 0
(C) ∞
(D) π/2
Solution:
Slope y – axis = tan 90° = ∞

Thus, option (c) is correct.

Question 4.
Line represented by equation x × 1/2 + y × √3/2 = 5 is in following form:
(A) Symmetrical form
(B) Slope form
(C) Intercept form
(D) Normal form
Solution:

Question 5.
Equation of line passsing through origin and parallel to line 3x – 4y = 7 is :
(A) 3x – 4y = 1
(B) 3x – 4y – 0
(C)4x – 3y = 1
(D) 3y – 4x = 0
Solution:
Equation of line parallel to straight line
3x – 4y = 7
3x – 4y = c
Since it passes through origin
∴ 3 (0) – 4 (0) = c
⇒ c = 0
Thus, Required equation is 3x – 4y = 0
Therefore, option (b) is correct.

Question 6.
It is the length of perpendicular drawn from origin to the line x + √3y = 1 then value of p is :

Solution:

⇒ x cos 60° + y sin 60° = 1/2
Comparing (i) by x cos α + y sin α = p
p = 1/2
Thus, option (B) is correct.

Question 7.
If lines y = mx + 5 and 3x + 5y = 8 are perpendicular to each other, then value of m is :

Solution:
y = mx + 5 …(i)
and 3x + 5y = 8
⇒ 5y = -3x + 8
⇒ y = −3/5x + 8 …(ii)
Lines (i) and (ii) are perpendicular to each other

Thus, option (a) is correct.

Question 8.
Equation of line passes through point (1, – 2) and perpendicular to line 3x – 4y + 7 = 0 will be :
(A) 4x + 3y – 2 = 0
(B) 4x + 3y + 2 = 0
(C) 4x – 3y + 2 = 0
(D) 4x – 3y – 2 = 0
Solution:
Equation of line perpendicular to
3x – 4y + 7 = 0
4x + 3y + c = 0
It passes through point (1,2)
4(1) + 3(-2) + c = 0
⇒ 4 – 6 + c = 0
⇒ c – 2=0
⇒ c = 2
Thus, equation is 4x + 3y + 2 =0
Thus, option (b) is correct.

Question 9.
Obtuse angle between lines y = -2 and y = x + 2 is :
(A) 145°
(B) 150°
(C) 135°
(D) 120°
Solution:
comparing y = -2x with y = m₁x + c₁
m₁ = 0
comparing y = x + 2 with y = m₂x + c₂
m₂ = 1
Angle between both lines

When tan θ = +1 then θ = 45°
when tan θ = -1 then θ = 135°
Thus, option (c) is correct

Question 10.
Length of intercepts cut at x and y axis by line 3x – 4y – 4 = 0 is :

Solution:

Thus, option (a) is correct

Question 11.
Line joining the points (1, 0) and (-2, √3 ) makes an angle θ with x-axis then value of tan θ is :

Solution:
Slope of line joining the points (1,0) and (-2, 3–√)
m = tanθ

Thus, option (d) is correct.

Question 12.
By reducing the equation of line 2x + √3 y – 4 = 0 in slope form, constant used in slope form is :

Solution:

Thus, option (d) is correct.

Question 13.
Find the equation of line which passes through point (2, 3) and makes an angle of 45° with x- axis.
Solution:
Given θ = 45°
then tan θ = m = tan 45° = 1
⇒ m = 1
equation of line passing through point (2, 3)
y – y₁ = m (x – x₁)
⇒ y – 3 = 1 (x – 2)
⇒ y – 3 = x – 2
⇒ x – y + 3 – 2 = 0
⇒ x – y + 1 = 0
Thus required equation of straight line is
x – y + 1 = 0

Question 14.
Find the equation of line which passes through points (-3, 2) and cut equal and opposite sign intercepts with axis.
Solution:
Let a and – a are intercepts cut by line, then its equation

Question 15.
If length of perpendicular from origin to straight line 4x + 3y + a = 0 is 2, then find value of a.
Solution:
Given equation
4x + 3y + a = 0
Length of perpendicular drawn from origin (0,0) to line 4x + 3y + a = 0

Question 16.
If intercepts cut at axis by any line bisects at point (5, 2), then find the equation of line.
Solution:
Let equation of line is x/a + y/b = 1 which meets x and y axis at points A(a, 0) and 5(0, b) coordinates of mid point of AB are

But it is given that mid point of AB is a
⇒ a/2 = 5
⇒ a = 10
⇒ b/2 = 2
⇒ b = 4
∴ Required equation
x/10 + y/4 = 1
⇒ 4x + 10y = 40
or 2x + 5y = 20

Question 17.
Find the equation of line which passes through a point (0,1) and intercept cut by line at axis as 3 times the intercept cut at y-axis.
Solution:
Let equation of line
x/10 + y/10 = 1 …(i)
Given that intercept of x-axis is 3 times the intercept of y-axis
a = 3b
Putting a = 3b in eq.ⁿ (i)
x/3b + y/b = 1
⇒ x + 3y = 3b
But it passes through point (0, 1)
∴ 0 + 3 (1) = 3b
⇒ b = 1
∴ a = 3 (1) = 3
∴ a = 3, b = 1
From eq.ⁿ (ii) x + 3y = 3 × 1
Thus, required equation of line
⇒ x + 3y = 3

Question 18.
If straight lines y = 2mx + c and 2x – y + 5 – 0 are parallel and perpendicular to each other, then find the value of m.
Solution:
Given lines
y = 2mx + c …(i)
and 2x – y + 5 = 0
⇒ y = 2x + 5 …(ii)
From equations (i) and (ii)
(i) lines will be II if
m₁ = m₂
⇒ 2m – 2
⇒ m = 1
(ii) lines will be perpendicular if
m₁m₂ = -1
⇒ 2m × 2 = -1
⇒ m = −1/4
Thus values of m will be 1 and −1/4

Question 19.
If length of perpendicular from origin to straight line 4x + 3y + a = 0 is 2, then find the value of a.
Solution:
Given equation 4x + 3y + a = 0
Length of perpendicular drawn from origin (0, 0) to line 4x + 3y + a = 0

Question 20.
If length of perpendicular from origin to line x/a + y/b = 1 is p, then prove that

Solution:
Equation of line cut intercepts a and b at axis
x/a + y/b = 1 …(i)
Length of perpendicular drawn from origin (0,0) to line

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