Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.18

Question 1: Find the equation of the straight lines passing through the origin and making an angle of 45° with the straight line √3x+y = 11.

Solution:

As line 1 passes through origin, there won’t be any intercept, and it will be in the form of y = mx (m as slope)

For line 2: √3x+y = 11, slope is M = -√3

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 45°, m=m and M = -√3

tan 45° = 

1 = 

We will have two cases,

1 =  and 1 = 

1-√3m = -√3-m and 1-√3m = √3+m

√3m-m = 1+√3 and √3m+m = 1-√3

m =  and m = 

By rationalizing, we get

m =  and m = 

m =  and m = 

m = 2+√3 and m = -(2-√3)

m = √3+2 and m = √3-2

Hence, the equation of line will be,

y = (√3+2)x and y = (√3-2)x

Question 2: Find the equation to the straight lines which pass through the origin and are inclines at an angle of 75° to the straight line x+y+√3(y-x)=a.

Solution:

As line 1 passes through origin, there won’t be any intercept and it will be in the form of y = mx (m as slope)

For line 2: x+y+√3(y-x)=a, 

x+y+√3y-√3x=a

x(1-√3)+y(1+√3)=a

slope is M = 

After rationalizing, we get

M =  = 2-√3

Here, it is given that these two lines make an angle of 75° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 75°, m=m and M = 2-√3

tan 75° = 

tan (45°+30°)= 

Using the trigonometric identity, 

We will have two cases,

2+√3 =  and 2+√3 = 

(2+√3)(1+m(2-√3)) = 2-√3-m and (2+√3)(1+m(2-√3)) = -(2-√3-m)

(2+√3+m(2²-(√3)²) = 2-√3-m and (2+√3+m(2²-(√3)²) = √3-2+m

2+√3+m(4-3) = 2-√3-m and 2+√3+m(4-3) = √3-2+m

2+√3+m+m = 2-√3 and 2+√3+m-m = √3-2

2m = -2-√3+2-√3 and 2+√3 = √3-2

2m = -2√3 and m is not defined

m = -√3 and m is not defined

Hence, the equation of line will be,

y = -√3x and x = 0

Question 3: Find the equation of the straight lines passing through (2,-1) and making an angle of 45° with the line 6x+5y-8=0.

Solution:

As line 1 passes through (2,-1), then it will be in the form of 

y-(-1) = m(x-2)  (m as slope)

y+1 = m(x-2)

For line 2: 6x+5y-8=0

5y = -6x + 8

slope is M = 

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 45°, m=m and M = 

tan 45° = 

We will have two cases,

1 =  and 1 = 

5-6m = -6-5m and 5-6m = -(-6-5m)

6m-5m = 5+6 and 5-6m = 5m+6

m = 11 and 5m+6m = 5-6

m = 11 and m = 

Hence, the equation of line will be,

y+1 = 11(x-2) and y+1 = 

11x-y-23=0 and 11y+x+9 = 0

Question 4: Find the equation to the straight lines which pass through the point (h,k) and are inclined at angle tan^-1 m to the straight line y=mx+c.

Solution:

As line 1 passes through (h,k), then it will be in the form of

y-(k) = M(x-h)  (M as slope)

For line 2: y=mx+c (m as slope)

Here, it is given that these two lines make an angle of tan^-1 m between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = tan^-1 m,

tan (tan^-1 m) = 

We will have two cases,

m =  and m = 

m(1+mM) = M-m and m(1+mM) = -(M-m)

m+m²M = M-m and m+m²M = m-M

2m = M-m²M and m²M = -M

M =  and M = 0

Hence, the equation of line will be,

y-(k) = 0(x-h) and y-k = 

y-k = 0 and (y-k)(1-m²) = (2m)(x-h)

Question 5: Find the equation to the straight lines passing through the point (2,3) and inclined at 45° to the line 3x+y-5=0.

Solution:

As line 1 passes through (2,3), then it will be in the form of

y-3 = M(x-2)  (M as slope)

For line 2: 3x+y-5=0

y = -3x+5

slope m = -3

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 45°,

tan 45° = 

We will have two cases,

 and 

1-3M = M+3 and 1-3M = -(M+3)

M+3M = 1-3 and 1-3M = -M-3

4M = -2 and 3M-M = 1+3

M =  and 2M = 4

M =  and M = 2

Hence, the equation of line will be,

 and y-3 = 2(x-2)

2y-6 = -(x-2) and y-3 = 2x-4

x+2y-8=0 and 2x-y-1=0

Question 6: Find the equation to the sides of an isosceles right-angled triangle the equation of whose hypotenuse is 3x+4y=4 and the opposite vertex is the point (2,2).

Solution:

As △ABC is an isosceles right angled triangle at B.

∠A = ∠C = 45°

We can say that, AB and BC makes 45° with AC.

Let the slope of AB as m₁ and BC as m₂.

AB: (y-2) = m₁(x-2)

BC: (y-2) = m₂(x-2)

Slope of AC : 3x+4y=4

Slope of AC is 

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 45°, and M = 

tan 45° = 

We will have two cases,

1 =  and 1 = 

4-3 = -3-4 and 4-3 = -(-3-4)

4m-3 = -3-4 and 4-3 = 3+4

m = -7 and 4+3 = 4-3

m = -7 and m = 

Hence, the equation of lines will be,

AB: (y-2) = m₁(x-2)

7y-x-12=0

BC: (y-2) = m₂(x-2)

y-2 = -7(x-2)

7x+y-16=0

Question 7: The equation of one side of an equilateral triangle is x-y=0 and one vertex is (2+√3,5). Prove that a second side is y+(2-√3)x=6 and find the equation of the third side.

Solution:

As equation of one side of equilateral triangle is x-y=0 and one vertex is (2+√3,5),

As, the point (2+√3,5) does not satisfy x-y=0. Then this is the vertex opposite of the line x-y=0

In equilateral triangle,

∠A = ∠B = ∠C = 60°

We can say that, AC and BC makes 60° with AB.

Let the slope of AC as m₁ and BC as m₂.

AB: (y-5) = m₁(x-(2+√3))

BC: (y-5) = m₂(x-(2+√3))

Slope of AC : x-y=0

y = x

Slope of AC is 1.

Here, it is given that these two lines make an angle of 60° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 60°, m = m₁ and m₂ and M = 1

tan 60° = 

√3= 

We will have two cases,

√3 =  and √3 = 

m =  and √3(1+m) = -(1-m)

m =  and √3+√3m = m-1

m =  and √3m-m = -1-√3

m =  and m = 

After rationalizing, we get

m =  and m = 

m =  and m = 

m = -(2-√3) and m = -(2+√3)

Hence, the equation of lines will be,

AB: (y-5) = m₁(x-(2+√3))

y-5 = -(2-√3)(x-(2+√3))

y-5 = -(2-√3)x+ (2²-(√3)²)

(2-√3)x+y-6 = 0

BC: (y-5) = m₂(x-(2+√3))

(y-5) = -(2+√3)(x-(2+√3))

y-5 = -(2+√3)x+(2+√3)^2

(2+√3)x+y-5 = 4+3+4√3

(2+√3)x+y = 12+4√3

Question 8: Find the equation of the two straight lines passing through (1,2) forming two sides of a square of which 4x+7y=12 is one diagonal.

Solution:

Let the point opposite of diagonal 4x+7y=12 be C(1,2)

Here, △BCD form an isosceles right angled triangle at C.

∠B = ∠D = 45°

We can say that, CD and BC makes 45° with BD.

Let the slope of CD as m₁ and BC as m₂.

CD: (y-2) = m₁(x-1)

BC: (y-2) = m₂(x-1)

Slope of BD : 4x+7y=12

Slope of BD is 

Here, it is given that these two lines make an angle of 45° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 45° and M = 

tan 45° = 

We will have two cases,

1 =  and 1 = 

7-4m = -4-7m and 7-4m = -(-4-7m)

7m-4m = -4-7 and 7-4m = 4+7m

3m = -11 and 7m+4m = 7-4

m =  and 11m = 3

m =  and m = 

Hence, the equation of lines will be,

CD: (y-2) = m₁(x-1)

3y-6 = -11x+11

11x+3y-17=0

BC: (y-2) = m₂(x-1)

11y-22 = 3x-3

3x-11y+19=0

Question 9: Find the equation of the two straight lines passing through (1,2) and making an angle of 60° with the line x+y=0. Find also the area of the triangle formed by the three lines.

Solution:

As equation of one side of equilateral triangle is x+y=0 and one vertex is (1,2),

As, the point (1,2) does not satisfy x+y=0. Then this is the vertex opposite of the line x+y=0

Hence, the lines make an equilateral triangle,

∠A = ∠B = ∠C = 60°

We can say that, AC and BC makes 60° with AB.

Let the slope of AC as m₁ and BC as m₂.

AB: (y-2) = m₁(x-1)

BC: (y-2) = m₂(x-1)

Slope of AC : x+y=0

y = -x

Slope of AC is -1.

Here, it is given that these two lines make an angle of 60° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 60° and M = -1

tan 60° = 

√3= 

We will have two cases,

√3 =  and √3 = 

√3(1-m) = -1-m and √3(1-m) = 1+m

√3-√3m = -1-m and √3-√3m = m+1

√3m-m = √3+1 and +√3m+m = √3-1

m =  and m = 

After rationalizing, we get

m =  and m = 

m =  and m = 

m = 2+√3 and m = 2-√3

Hence, the equation of lines will be,

AB: (y-2) = (2+√3)(x-1)

y-2=(2+√3)x-2-√3

(2+√3)x-y-√3=0 ……………….(i)

BC: (y-2) = (2-√3)(x-1)

y-2=(2-√3)x-2+√3

(2-√3)x-y+√3=0 ……………….(ii)

Using (i) and x+y=0, we get

C = (1,2)

Using distance formula, AC

Area of equilateral triangle ABC,

 sq. unit

Question 10: Two sides of an isosceles triangle are given by the equations 7x-y+3=0 and x+y-3=0 and its third side passes through the point (1,-10). Determine the equation of the third side.

Solution:

Let the equation of the line AB and AC 7x-y+3=0 and x+y-3=0 respectively.

∠B = ∠C

Slope of line AB: 7x-y+3=0

y = 7x+3

Slope m₁ = 7

Slope of line AC: x+y-3=0

y = -x+3

Slope m₂ = -1

Let the slope of line BC be m, which passes through the point (1,-10)

y-(-10) = m(x-1)

y+10 = m(x-1)

The angle between the lines AB and BC is equal to the angle between the lines AC and BC, say θ

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

By taking positive sign

tan θ = 

tan θ = 

Solving it, we get

m = -3 or 

By taking negative sign

tan θ = 

tan θ = 

Solving it, we get

m² = -1 (which is not possible)

Line BC: when m = -3

y+10 = -3(x-1)

y+10 = -3x+3

3x+y-13=0

Line BC: when m = 

3y+30 = x-1

x-3y-31=0

Question 11: Show that the point (3,-5) lies between the parallel lines 2x+3y-7=0 and 2x+3y+12=0 and find the equation of lines through (3,-5) cutting the above lines at an angle of 45°.

Solution:

Let the line 1: 2x+3y-7=0

Line 2: 2x+3y+12=0

As the slope of line 1 and line 2 is same, these lines are parallel.

Slope of lines M = 

To check whether (3,-5) lies between these lines

Taking x=3 and y=-5

(2x+3y-7)(2x+3y+12)<0 (If its true then the point lies between these lines)

(2(3)+3(-5)-7)(2(3)+3(-5)+12)

(6-15-7)(6-15+12)

(-16)(3) <0, which is a negative value.

Hence, the point (3,-5) lies between the lines.

Let the slope of the line be m, which is passing through the point (3,-5)

y-(-5) = m(x-3)

y+5 = m(x-3)

As it is given that the line is cutting above lines at an angle of 45°.

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Taking θ = 45° and M = 

By taking positive sign

tan 45° = 

We will have two cases,

1 =  and 1 = 

3-2m = -2-3m and 3-2m = -(-2-3m)

3m-2m = -2-3 and 3-2m = 2+3m

m = -5 and 3m+2m = 3-2

m = -5 and m = 

Hence, the equation of line will be,

when m = -5

y+5 = (-5)(x-3)

y+5 = -5x+15

5x+y-10=0

when m = 

y+5 = (x-3)

5y+25 = x-3

x-5y-28=0

Question 12: The equation of the base of an equilateral triangle is x+y=2 and its vertex is (2,-1). Find the length and equations of its sides.

Solution:

∠A = ∠B = ∠C = 60°

We can say that, AC and AB makes 60° with BC.

Let the slope of AC as m₁ and AB as m₂ which passes through the point (2,-1).

AB: (y-(-1)) = m₁(x-2)

y+1 = m₁(x-2)

AC: (y-(-1)) = m₂(x-2)

y+1 = m₂(x-2)

Slope of BC : x+y=2

y = -x+2

Slope of BC is -1.

Here, it is given that these two lines make an angle of 60° between them. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 60°, m = m₁ and m₂ and M = -1

tan 60° = 

√3= 

We will have two cases,

√3 =  and √3 = 

√3(1-m) = -1-m and √3(1-m) = 1+m

√3-√3m = -1-m and √3-√3m = m+1

√3m-m = √3+1 and +√3m+m = √3-1

m =  and m = 

After rationalizing, we get

m =  and m = 

m =  and m = 

m = 2+√3 and m = 2-√3

Hence, the equation of lines will be,

AB: y+1 = m₁(x-2)

y+1 = (2+√3)(x-2)

y+1 = (2+√3)x-4-2√3

(2+√3)x-y-5-2√3=0 ……………….(i)

AC: y+1 = m₂(x-2)

y+1 = (2-√3)(x-2)

y+1 = (2-√3)x-4+2√3

(2-√3)x-y-5+2√3=0 ……………….(ii)

Using (ii) and x+y=2, we get

A = (2,-1)

Using distance formula, AC

AC = AB = BC = 

Question 13: If two opposite vertices of a square are (1,2) and (5,8), find the coordinates of its other two vertices and the equations of its sides.

Solution:

Let’s consider a square ABCD.

According to the property of square, the diagonal bisects the angle.

Hence, ∠AOB = ∠AOD = 45°

Slope of the diagonal AC:

Let the slope of the line AB and AD be m1 and m2, which passes through the point (1,2)

AB: y-2 = m₁(x-1)

AD: y-2 = m₂(x-1)

Here, it is given that these two lines make an angle of 45° with AC. Hence,

As we know that, two lines having slopes as m and M will have angle θ between them as follows:

Here, we have θ = 45°, m = m₁ and m₂ and M = 

tan 45° = 

We will have two cases,

1 =  and 1 = 

2+3m = 3-2m and 2+3m = -(3-2m)

3m+2m = 3-2 and 2+3m = 2m-3

5m = 1 and 3m-2m = -3-2

m =  and m = -5

m =  or -5

Hence, the equation of lines will be,

AB: y-2 = m₁(x-1)

5y-10 = x-1

x-5y+9=0 ……………….(i)

AD: y-2 = m₂(x-1)

y-2 = -5(x-1)

y-2 = -5x+5

5x+y-7=0 ……………….(ii)

As BC is parallel to AD, So 

The equation BC will be 5x+y+λ=0 and as BC passes through C(5,8), we get

5(5)+(8)+λ=0

33+λ=0

λ = -33

Hence, Equation of BC is 5x+y-33=0 ……………….(iii)

Now as CD is parallel to AB, So

The equation CD will be x-5y+λ=0 and as CD passes through C(5,8), we get

5-5(8)+λ=0

-35+λ=0

λ = 35

Hence, Equation of CD is x-5y+35=0 ………………………(iv)

Solving (i) and (iii), we get

B(6,3)

Solving (ii) and (iv), we get

D(0,7)

Hence, the equation of lines are

AB = x-5y+9=0

BC = 5x+y-33=0

CD = x-5y+35=0

AD = 5x+y-7=0

And the vertices of square are

A(1,2), B(6,3), C(5,8) and D(6,3)