Math Assignment Class XI Ch -09

Math Assignment Class XI Ch - 09

Extra questions of chapter 09 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

MATHEMATICS ASSIGNMENT ON

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 45^oto 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)

⇒ 5x₁- y₁+ 2 = 0 ⇒ y_1
=5x₁+ 2 ...... (iii)

Eqn. (i) passes through (x₂, y₂)

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

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

1/2(x₁+ x₂) = 2 ⇒ x₁+ x₂ = 4 ....... (v)

1/2(y₁+ y₂) = 4 ⇒ y₁+ y₂ = 8 ...... (vi)

Putting the value of y₁ 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

x₁= 1, x₂= 3

Putting x₁= 1, in eqn.(iii) we get

y_{1 =}5 ✕ 1+ 2 = 7

⇒ (x₁, y₁) = (1, 7)

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

3x + y - 10 = 0

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
AB is the line segment of the required line between the given line.
Point P is the mid-point of AB.
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 previous question we find the coordinates of point A as (5, 9)
Find the slope of AB and then equation of AB 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

x₁= 2, 4

Putting the values in eqn(i) we get

y₁= 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: Equation of line is √3x - y + 3 - 2√3 = 0

Solution Hint

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

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 tan60^o 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(x - x₁)

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 = 45^o 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 = 90^oand AB = BC so ∠ACB = 45^o

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

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

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.

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