### CBSE Assignments class 09 Mathematics

Mathematics Assignments & Worksheets  For  Class IX Chapter-wise mathematics assignment for class 09. Important and useful extra questions strictly according to the CBSE syllabus and pattern with answer key CBSE Mathematics is a very good platform for the students and is contain the assignments for the students from 9 th  to 12 th  standard.  Here students can find very useful content which is very helpful to handle final examinations effectively.  For better understanding of the topic students should revise NCERT book with all examples and then start solving the chapter-wise assignments.  These assignments cover all the topics and are strictly according to the CBSE syllabus.  With the help of these assignments students can easily achieve the examination level and  can reach at the maximum height. Class 09 Mathematics    Assignment Case Study Based Questions Class IX

## Math Assignment  Class XI  Ch - 10STRAIGHT LINES

Extra questions of chapter 10 Straight Lines class 11  with answers and  hints to the difficult questions, strictly according to the CBSE & DAV Board syllabus. Important and useful math. assignment for the students of class 11

## STRAIGHT LINES (XI)

Strictly according to the CBSE and DAV Board

Question 1

Two lines passing through point (2, 3) are inclined at an angle of 45to each other. If the slope of one of the line is 2, than find the slope and equation of the another line.

Solution Hint

$tan(45^{o})=\left|\frac{m-2}{1+2m}\right|$

$\pm 1=\frac{m-2}{1+2m}$

Taking +ve sign we get the value of m = 1/3

Then eqn. of line is  x - 3y + 7 = 0

Taking -ve sign we get the value of m = -3

Then eqn. of line is  3x + y - 9 = 0

Question 2

Find the equation of the line passing through the point of intersection of the lines x - y + 1 = 0 and 2x - 3y + 5 = 0 and whose distance from the point (3, 2) is 7/5 units

Ans:  4x - 3y + 1 = 0  or  3x - 4y + 6 = 0

Solution

Given that required line passes through the intersection of both the lines equation of required line is (x - y + 1) + k (2x - 3y + 5) = 0, for some value of k

(1 + 2k)x + (-1 - 3k)y + (1 + 5k) = 0 ...... (i)

Distance of the required line from the point (3, 2) is 7/5

So by using distance formula

$d=\frac{\left| ax_{1}+by_{1}+c\right|}{\sqrt{a^{2}+b^{2}}}$

we get the value of k = -1/6 and 1

Putting k= -1/6 in eqn.(i) we get the eqn. of line

4x - 3y + 1 = 0

Putting k=1 in eqn.(i) we get

3x - 4y + 6 = 0

Question 3

A line is such that its segment between the lines 5x - y + 2 = 0 and x + 3y  - 6 = 0 is bisected at the point (2, 4). Obtain its equation.

Ans: 3x + y =10

Solution

Given equations of lines are:
5x - y + 2 = 0 ......... (i)
x + 3y  - 6 = 0 ........ (ii)

Eqn. (i) passes through (x1,y1)

5x- y+ 2 = 0     y1 = 5x1  + 2 ...... (iii)

Eqn. (i) passes through (x2, y2)

x2 + 3y2  - 6 = 0  y= 1/3(6 - x2) ..... (iv)

Also the mid point of the required line is (2, 4)

1/2(x+ x2) = 2 x+ x2 = 4  ....... (v)

1/2(y+ y2) = 4  y+ y2 = 8  ...... (vi)

Putting the value of y and y2  from equation(iii) and (iv) in eqn(vi)

5x1  + 2 + 1/3(6 - x2) = 8

Solving this equation we get

15x1 - x2 = 12  ...... (vii)

Solving eqn.(v) and (vii) we get

x= 1,  x= 3

Putting x= 1, in eqn.(iii)  we get

y1 = 51 + 2  = 7

(x1,y1) = (1,7)

Equation of line passing through (1, 7) and (2, 4)

3x + y - 10

Question 4

A line is such that its segment between the lines 4x + 3y – 21 = 0 and 10x + y – 59 = 0 is bisected at the point (4, 6). Find its equation.

Ans: 3x - y - 6 = 0

Solution Hint

Above problem can be represented as shown in the figure given below

• Let required line is intersected by the given two lines at points A and B respectively
• AC is the line segment of the required line between the given line.
• Point P is the mid-point of AC.
• Line 10x + y – 59 = 0 Passes through the Point A and line 4x + 3y – 21 = 0 passes through the point B.
• Solving these equations according to the question we find the coordinates of point A as (3, 3)
• Find the slope of AC and then equation of AC becomes 3x - y - 6 = 0
Question 5
If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y= mx + 4, find the value of m.
Ans:
$m=\frac{1\pm 5\sqrt{2}}{7}$
Solution
Find the slopes of the given lines
Find tan θ in both the cases and comparing these two for the values of m
Rejecting m2 = -1, because square of any number never be -ve

Question 6

The points (1, 3) and (5, 1) are two opposite vertices of a rectangle. The other two vertices lie on the line y = 2x + c. Find c and the remaining vertices.

Ans: c = -4

Other two vertices are (2,0) and (4,4)

Solution Hint

Given situation can be represented as shown in the figure

Since diagonals of a rectangle are equal and bisect each other

So mid point of AC = mid point of BD

Coordinates of mid point of AC = (3, 2)

This point lie on the line y = 2x + c

⇒ 2 = 2 x 3 + c  ⇒  c = - 4

Now equation of BD is y = 2x - 4

Since BD Passes through B(x1, y1) so we have

y1 = 2x1 - 4 ...... (i)

Since AB BC

So slope of AB x BC = -1

$\frac{y_{1}-3}{x_{1}-1}\times \frac{y_{1}-1}{x_{1}-5}=-1$

Solving this we get

$y_{1}^{2}-4y_{1}+x_{1}^{2}-6x_{1}+8=0$

Now putting eqn(i) in this equation we get
x= 2, 4
Putting the values in eqn(i) we get
y= 0, 4
Required vertices are (2, 0) and (4, 4)

Question 7

If p and q are the lengths of perpendicular from the origin to the lines x cosθ - y sin θ = k cos 2θ and x sec θ + y cosec θ = k respectively. Prove that  P2 + 4q2 = k2

Solution Hint:

Find the values of P and Q by using formula to finding the perpendicular distance of the line from the (0,0)

Now putting the value of P and Q in the LHS of P2 + 4q2 .

Question 8

A vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is √3x + y = 2. Find the equation of the other two sides.

Ans:

Solution Hint

Here ABC is an equilateral triangle so its each angle = 60o

Let A (2, 3) be a vertex of an equilateral triangle ABC,

then the equation of the side BC is √3x + y = 2

Slope of BC = - √3

Let slope of AB = m

Now find tan60o by using the formula

$tan(60^{o})=\frac{\left|m-(-\sqrt{3}) \right|}{1+\sqrt{3}(-m)}$

$\pm \sqrt{3}=\frac{m+\sqrt{3} }{1-\sqrt{3}m}$

Taking positive sign and find the value of m = 0

Equation of line is y - 3 = 0

Taking - ve sign and find the value of  m =√3

Equation of line is √3x - y + 3 - 2√3 = 0

Question 9
Find the length of the perpendicular from the origin to the line joining the two points whose coordinates are A(cos α, Sin α) and B(cos β, sin β).
Ans:
$d=\left| cos\left ( \frac{\alpha -\beta }{2} \right )\right|$

Solution Hint:
Find the equation of line passing through the two points A and B by using

y - y1 = m(y - y1)

Simplify the equation by using the AB formulas and cos(A + B)

Now find the ⊥ distance of this line from the origin we get

$d=\left| cos\left ( \frac{\alpha -\beta }{2} \right )\right|$

Question 10

Ram and Rahim are cousins living in the same colony. The lanes of their houses are represented by the lines 4x - 3y + 6 = 0 and 4x - 3y - 9 = 0, respectively.

Based on the above information, answer the following questions:
(i)Find the slope of the lane where Rahim lives?
(ii)Find the angle between the two lanes?
(iii) What is the distance of the point (0, 2 ) in Ram’s lane to Rahim’s lane?

Ans: (i) Slope = 4/3

(ii) Lines are parallel so angle between then is 0

(iii) Distance of the line from the point is 3

Question 11

A policeman was standing at the junction (crossing) of two straight paths represented by the equations 2x + 3y + 4 = 0 and 3x + 4y - 5 = 0. He saw a thief running along the path 6x - 7y + 8 = 0. The policeman ran and caught the thief when he was at the shortest distance from the junction. Find the equation of path followed by policeman.

Ans: 7x +6y - 85 = 0

Solution Hint:

Above problem can be represented as shown in the figure given below

Solve the equation AB and CD and find the value of x and y.

Coordinates of the point P where the policeman standing are (31, -22)

Slope of equation  EF is 6/7

PQ is ⊥ to EF because shortest distance is always perpendicular.

Slope of PQ = -7/6

Now find the equation of PQ

Question 12

One side of a rectangle lies along the line 4x + 7y + 5 = 0 two of its opposite vertices are (- 3, 1) and (1,1) find the equations of the other three sides

Ans: 4x + 7y - 11 = 0,  7x - 4y - 3 = 0, 7x - 4y + 25 = 0

Solution Hint:

Above problem can be represented as shown in the figure given below

AB || CD so eqn. of CD will be
4x + 7y + k = 0

CD passes through (1, 1) ⇒ k = -11

So equation of CD is 4x + 7y - 11 = 0

AB ⊥ BC so Eqn. of BC will be 7x - 4y + k = 0

BC passes through (1, 1)  ⇒ k = -3

So equation of BC is = 7x - 4y - 3 = 0

AD || BC so eqn. of AD will be 7x - 4y + k = 0

AD passes through (-3, 1) ⇒ k = 25

So equation of AD is = 7x - 4y + 25 = 0

Question 13

The hypotenuse of isosceles right triangle lies along the line 2x - y = 4 and vertex opposite to hypotenuse is (1, 5),  obtain the equation of other side.

Ans: Eqn. of AC :   3x + y - 8 = 0  or x - 3y - 14 = 0

Eqn. of BC : x - 3y + 14 = 0 or  3x + y - 8 = 0

Solution Hint

Above problem can be represented as shown in the figure given below

Slope of AB = 2

Let Slope of AC = m

Angle between AB and AC = 45o so by using  $tan45^{o} =\frac{|m-2|}{1+2m}$   find the value of m

By taking + ve sign we get m = - 3

Eqn. of AC is 3x + y - 8 = 0

Since BC is ⊥ to AC so eqn. of BC will be x - 3y + k = 0

Since BC passes through the point (1, 5) so we get k = 14

Eqn. of BC is x - 3y + 14 = 0

Now comparing - ve sign we get the value of m = 1/3

Eqn. of AC become: x - 3y - 14 = 0

Since BC is ⊥ to AC so eqn. of BC will be 3x + y + k = 0

Since BC passes through the point (1, 5) so we get k = -8

Eqn. of BC is  3x + y - 8 = 0

Question 14

If one diagonal of a square is along the line 8x – 15y = 0 and one of its vertex is at (1, 2) then find the equations of sides of square passing through this vertex.

Ans: Equation of one side is : 23x - 7y - 9 = 0

Equations of another side is : 7x + 23y - 53 = 0

Solution Hint:

Above problem can be represented as shown in the figure given below

Slope of diagonal AC = 8/15

Let slope of BC = m

In ABC, B = 90and AB = BC so ACB = 45o

$tan45^{o}=\frac{|m-\frac{8}{15}|}{1+m\times \frac{8}{15}}$

$\Rightarrow \frac{15m-8}{15+8m}=\mp 1$

Taking +ve sign we get  slope m = 23/7

Equation of BC becomes 23x - 7y - 9 = 0

Taking -ve sign we get slope m = -7/23

We get the equation of AB because AB ⊥ BC

Equation of AB is 7x + 23y - 53 = 0

Question 15

The image of a point with respect to the line 2x – y + 6 = 0, assuming the line to be a mirror, is (6, 6). Find the point. Also find the equation of line joining this point and its image.

Ans: Coordinates of required point are (-18/5, 54/5)

Required line CD is given by x + 2y - 18 = 0

Solution Hint

• Let given eqn. of line  is represented by AB
• Let required point is C(x, y)
• Image of of point C in the line AB is D(6, 6)
• So according to the question AB is the ⊥ bisector of CD and point P is the mid point of CD.
• Find the equation of CD which is x + 2y - 18 = 0 and then solve AB and CD we get coordinates of point P(6/5, 42/5)
• Point P is the mid point of AB and CD so by using mid point formula we get  point C (-18/5, 54/5) respectively.