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### Math Assignment Class XI Ch -10 | Straight Lines

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## Math Assignment Class XI Ch - 10

**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**

**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**

^{o }to each other. If the slope of one of the line is 2, than find the slope and equation of the another line.

**Solution Hint**

**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 **

**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 (x _{1},y_{1})**

**⇒ 5x _{1 }-
y_{1 }+ 2 = 0 ⇒ y_{1
= }5x_{1 }+ 2 ...... (iii)**

**Eqn. (i) passes through (x _{2}, y_{2})**

**x _{2} + 3y_{2} - 6 =
0 ⇒ y_{2 }= 1/3(6
- x_{2}) ..... (iv)**

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

**1/2(x _{1 }+ x_{2}) = 2 ⇒ x_{1 }+ x_{2} =
4 ....... (v)**

**1/2(y _{1 }+ y_{2}) =
4 ⇒ y_{1 }+ y_{2} =
8 ...... (vi)**

**Putting the value of y _{1 } and y_{2}
from equation(iii) and (iv) in eqn(vi)**

**5x _{1 }+ 2 + 1/3(6 - x_{2})
= 8**

**Solving this equation we get**

**15x _{1} - x_{2} =
12 ...... (vii)**

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

**x _{1 }= 1, x_{2 }=
3**

**Putting x _{1 }= 1, in eqn.(iii)
we get **

**y _{1 = }5 ✕ 1_{ }+ 2 = 7 **

**⇒ (x _{1}, y_{1})
= (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****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:**

**Solution**

**Find the slopes of the given lines**

**Find tan Î¸ in both the cases and comparing these two for the values of m**

**Rejecting m**

^{2}= -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(x _{1}, y_{1}) so we have**

**y _{1} = 2x_{1} - 4 ...... (i)**

**Since AB ⊥ BC **

**So slope of AB x BC = -1**

**Solving this we get**

**Now putting eqn(i) in this equation we get**

**x**

_{1 }= 2, 4**Putting the values in eqn(i) we get**

**y**

_{1 }= 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 cosÎ¸ - y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k respectively.
Prove that P ^{2} + 4q^{2} = k^{2} **

**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 P ^{2} + 4q^{2} .**

**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**

**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 Î±, Sin Î±) and B(cos Î², sin Î²).**

**Ans:**

**Solution Hint:**

**Find the equation of line passing through the two points A and B by using**

**y - y _{1} = m(x - x_{1})**

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

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

** **

**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

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 **

**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 ^{o }and AB = BC so ∠ACB = 45^{o}**

**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.**

**THANKS FOR YOUR VISIT**

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