# Class 11 RD Sharma Solutions – Chapter 23 The Straight Lines- Exercise 23.13

**Question 1: Find the angles between each of the following pairs of straight lines.**

**(i) 3x+y+12=0 and x+2y-1=0**

**Solution:**

Given equations of lines are,3x + y + 12 = 0, x + 2y -1 = 0

Letm

_{1}andm_{2}be the slopes of these lines respectively.By

y = mx +c,we getm_{1}=-3 and m_{2}=-1/2Let θ be the angle between the two lines,

By using formula

⇒

⇒ 1

Therefore,

The angles between the two lines is

45°.

**(ii) 3x-y+5 = 0 and x-3y+1 = 0**

**Solution:**

Given equations of lines are 3x – y + 5 = 0, x – 3y +1 = 0

Let m

_{1}and m_{2}be the slopes of these lines respectively.By y = mx +c, we get m

_{1}=3 and m_{2}=1/3Let θ be the angle between the two lines,

We know that,

⇒

Therefore,

The angle between the two lines is

**(iii) 3x+4y -7 = 0 and 4x-3y+5 = 0**

**Solution:**

Given equations of lines are 3x + 4y – 7 = 0, 4x – 3y+5 = 0

Letm

_{1}andm_{2}be the slopes of these lines respectively.By y= mx +c, we get m

_{1 }= and m_{2 }=Here, if we carefully observe, m

_{1}m_{2 }= -1,which meansFrom the formula

,denominator will become 0,Therefore, ,

The angle between the two lines is 90°.

**(iv) x-4y = 3 and 6x-y = 11**

**Solution:**

Given equations of lines are x – 4y =3, 6x – y =11

Letm

_{1}andm_{2}be the slopes of these lines respectively.By y = mx +c, we get

,m_{1}=1/4and m_{2}=6Let θ be the angle between the two lines,

We know that,

⇒

⇒

⇒

Therefore,

The angle between the two lines is

**(v) (m**^{2}-mn)y = (mn+n^{2})x + n^{3} and (mn+m^{2})y = (mn-n^{2})x + m^{3}

^{2}-mn)y = (mn+n

^{2})x + n

^{3}and (mn+m

^{2})y = (mn-n

^{2})x + m

^{3}

**Solution:**

Given two lines, letm

_{1}andm_{2}be the slopes of these lines.By y = mx +c, we getm

_{1 }= and m_{2 }=Let θ be the angle between two lines,

We know that

⇒

⇒

⇒

Therefore, Angle between two lines is.

**Question 2: Find the acute angle between the lines 2x-y+3 = 0 and x+y+2 = 0**

**Solution:**

Letm

_{1}andm_{2}be the slopes of these two linesBy y = mx +c, we get m

_{1}=2 and m_{2}=-1Let θ be the angle between the two lines,

We know that,

⇒

⇒

Here we need acute angle,is positive if the angle is acute and negative if obtuse.

Therefore,.

The acute angle between the two lines is.

**Question 3: Prove that points (2, -1), (0, 2), (2, 3), and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.**

**Solution:**

Let the given points are A = (2,-1), B = (0, 2), C = (2, 3) and D = (4, 0) are coordinates of a parallelogram.

For these points to form a parallelogram, it is must that any pair of two lines formed by these points are parallel to each other.So, Now lets find the slopes of lines AB, BC, CD, DA using formula

Slope of line

Slope of line

Slope of line

Slope of line

Since the lines

AB parallel to CD and BC parallel to DA, the points form a parallelogram.Now,

Angle between the diagonals of parallelogram = Angle between the lines AC and BD.Letm

_{1}andm_{2}be the slopes of these lines,From the figure, the angle between diagonals

We know that

⇒

Therefore, The angle between the diagonals is

**Question 4: Find the angles between the line joining the points (2, 0), (0, 3)**,** and the line x + y = 1.**

**Solution:**

Let slope of line joining the points (2, 0) and (0, 3) is m

_{1 }= -3/2slope of line m

_{2}=-1Let θ be the angle between the two lines,

We know that,

⇒

⇒

Therefore, the acute angle between the line and the line joining the points is

**Question 5: If **θ** is the angle which the straight line joining the points (x1,y1)and (x2,y2) subtends at origin, prove that****and**

**Solution:**

Let the points

A = (x_{1}, y_{1}), B = (x_{2}, y_{2}) and origin O = (0, 0)Slopes of lines joining OA and OB are m

_{1 }= y_{1}/x_{1}and m_{2}= y_{2}/x_{2}Let θ be the angle between the lines OA and OB.

We know that,

⇒

Therefore,

By the formula,we get

Substituting Tanθ from above equation, we get,

Therefore, hence proved.

**Question 6: Prove that the straight lines(a+ b)x+(a – b)y=2ab, (a – b)x+(a + b)y=2ab and x + y=0 form an isosceles triangle whose vertical angle is**

**Solution:**

Let m1, m2, m3 be the slopes of given lines respectively.

m

_{1 }= , m_{2 }= and m_{3 }= -1Let

θbe the angles between the lines_{1}, θ_{2}, θ_{3}Now,=

⇒

⇒⇒

We know that, using this above equation

⇒

=

⇒

⇒

=

⇒

⇒

Here, angle θ

_{2}and θ_{3}are equal, and θ_{1}is the vertical angleTherefore, the given lines forms Isosceles triangle with vertical angle

**Question 7: Find the angle between the lines x = a, by + c = 0**

**Solution:**

The given lines are in the form of

x = constantandy=constantrespectivelywhere x=c and y= -c/b

x = c line is parallel to y-axis as there is no

y-coefficientand is parallel to x-axis as there is no

x-coefficientTherefore, the Angle between the two lines is 90°

**Question 8: Find the tangent of **the **angle between the lines which have intercepts 3, 4**,** and 1, 8 on the axes respectively.**

**Solution:**

Equation of line which have intercepts a, b on x and y-axis is

Therefore, the line with intercepts 3,4 is

and the line with intercepts 1, 8 is

letm

_{1}and m_{2}be the slopes of these lines,m

_{1 }= and m_{2 }= -8Now, let θ be the angle between the lines,

⇒

⇒ ⇒

Therefore,

The tangent of angle between the lines is 4/7

**Question 9: Show that line a**^{2}x+ay+1 = 0 is perpendicular to line x-ay = 1

^{2}x+ay+1 = 0 is perpendicular to line x-ay = 1

**Solution:**

Letm

_{1}and m_{2}be the slopes of given lines,m_{1}=-a andm_{2}=1/aHere, if we carefully observe,m

_{1}m_{2}=-1,which meansFrom the formula,

denominator will become 0,Therefore,, .

The angle between the two lines is 90°, and they are

perpendicularto each other.Therefore, Hence proved.

**Question 10: Show that tangent of an angle between the lines****and**** is****.**

**Solution:**

Given lines,⇒

⇒

Let slopes of these lines arem

_{1}andm_{2}respectively.and

Now, let θ be the angle between the lines,

⇒

⇒

Therefore, Tangent of angle between the lines is

Hence, proved.