Important Questions Class 11 Maths Chapter 10

Important Questions Class 11 Mathematics Chapter 10

Important Questions for CBSE Class 11 Mathematics Chapter 10 – Straight Lines

Extramarks’ Important Questions Class 11 Mathematics Chapter 10, contains comprehensive questions that cover the main topics of Chapter 10 Straight Lines in Class 11 Mathematics. This set of questions is prepared by subject matter experts. Students can review these notes to gain a better understanding of the concepts, terminologies, and various practical problems in the chapter.

These Chapter 10 Class 11 Mathematics Important Questions will help students quickly revise the main concepts through a question-answer format. Students can directly access updated questions that are in accordance with the latest CBSE Syllabus. This set includes questions with solutions along with their marks’ distribution.

CBSE Class 11 Mathematics Chapter-10 Important Questions – Free Download

Study Important Questions for Class 11 Mathematics Chapter 10 – Straight Lines

Given below are a few of the Class 11 Mathematics Chapter 10 Important Questions. For the complete set, students can access the link provided on the Extramarks website.

Very Short Answers and Questions                                                                                                     

[1 or 2 Marks]

Q1. Find the equation of the line, which makes intercepts – 3 and 2 on the x and y axis respectively. 

Ans. Given,

        x-intercept a = -3.

y-intercept b = 2.

The required equation is given by xa+yb=1

a=-3,  b=2

∴ x-3+y2=1

2x-3y+6=0

Q2.  Find the value of x for which the points x,-1, 2,1, and 4,5 are collinear.

Ans. Given three collinear points.

Let the point be Ax,-1, B2,1 and C4,5

Since the points are collinear. Therefore,

Slope of AB= Slope of BC.

1+12-x=5-14-2

22-x=42

2-x=1

x=1

Q3. Find the value of k , given that the distance of the point4,1 from the line 3x-4y+k=0 is 4 units. 

Ans. Given,

The distance of the point 4,1 from the line 3x-4y+k=0 is 4 units.

3(4)-4(1)-k32   +-42 =4

12-4+k25=4

8+k5=4

8+k=20

When

8+k=20

k=12

When

-(8+k)=20

k=-28

Q4. Find the distance of a point -2,3 from the line 12x-5y=2.

Ans. Given the point -2,3 and the equation of the line 12x-5y=2.

Since we know the distance of a point x1,y1 from the line ax+by+c=0 is 

ax1+by1+ca2+b2

Distance of the point -2,3 from the line ax+by+c=0 is 12(-2)-5(3)-2(12)2   +-52

=-24-15-213=4113

Q5. Find the equation of a line whose perpendicular distance from the origin is 5 units and the angle between the positive direction of the x-axis and the perpendicular is 30°.

Ans. Given,

The perpendicular distance of the line from the origin is 5 units.

The angle between the positive direction of the  x-axis and the perpendicular is 30°.

Hence,

p=5, and =30°

The required equation is given by x cos +y sin =p

x cos 30°+y sin 30°=5

3x+y-10=0

Q6. Find x so that the inclination of the line containing the points x,-3, and (2,5) is 135° .

Ans. Imagine a line joining the points x,-3, and (2,5) which makes an angle of 135° with the x-axis .

We know that the slope of a line joining the two points (x1,y1) and (x2,y2) is equal to the tangent of the angle made by the line with x-axis in the anti-clockwise direction given as follows: slope

=tan=y2-y1x2-x1

So we have =135°, x1=2,   x2=x, y1=5, and y2=-3

tan 135°=-3-5x-2

Since we know that

tan 135°=tan(90°+45°)=-cot 45°=-1 as in the second quadrant, the tangent function is negative.

-1=-3-5x-2

-1=-8x-2

On cross-multiplication, we get as follows:

-x+2=-8

On adding -2 on both sides of the equation, we get as follows:

-x-2+2=-8-2

-x=-10

x=10

The value of x is equal to 10.

 Short Answers and Questions 

[4  Marks]

Q1. The slope of a line is double the slope of another line. If the tangent of the angle between them is 13, find the slopes of the lines.

Ans. Given,

The slope of a line is double the slope of another line.

Tangent of the angle between them,

I.e., tan =13

Let the slope of one line is m and the other line is 2m.

13=2m-m1+(2m)(m)

13=m1+2m2

13=m1+2m2

2m2-3m+1=0

2m2-2m-m+1=0

2m(m-1)-1(m-1)=0

(m-1)(2m-1)=0

m=1, m=12

-13=m1+2m2

2m2-3m+1=0

2m2-2m-m+1=0

2m(m-1)-1(m-1)=0

(2m-1)(m-1)=0

m=12 or m=1

-13=m1+2m2

-1-2m2=3m

2m2+3m+1=0

2m2+2m+m+1=0

2m(m+1)+1(m+1)=0

(2m+1)(m+1)=0

m=-1 or m=-12

Q2. The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14 per litre and 1220 litre of milk each week at Rs 16 per litre. Assuming a linear relationship between the selling price and demand, how many litres could he sell weekly at Rs 17 per litre?

Ans. Let the selling price be P along x-axis.

Let the demand for milk be D along y-axis.

We know that the equation of a line is

y=mx+c

Here, P is along x-axis and D is along y-axis.

So, our equation becomes

D=mP+c

Now,

Owner sells 980 litre milk at Rs 14 /litre

So, D=980 & P=14 satisfies the equation

Putting values in (1)

1220=16m+c

Therefore, the equations are

980=14m+c

1220=16m+c

From (A)

980=14m+c

980-14m=c

Putting the value of c in (B)

1220=16m+980-14m

1220-980=16m-14m

240=2m

m=120

Putting m=120 in (A)

980=14m+c

980-14m=c

980-14(120)=c

980-1680=c

-700=c

Putting value of m&c in (1)

D=mP+c

D=120P+700

the equation is D=120P+700

We need to find how many litres could he sell weekly at Rs 17/litre i.e. we need to find

D when P=17.

Putting P=17 in the equation.

D=120P+700

D=120(17)+700

D= 2040+700

D=1340

Hence when the price is Rs 17/litre , 1340 litres of milk could be sold.

Q3. Find the equations of the lines, which cut off intercepts on the axes whose sum and products are 1 and -6 respectively. [Mathematics Class 11 Chapter 10 Important Questions]

Ans. Let the intercepts made by the line on the axes be,

x-intercept=a

y-intercept=b

Sum of intercepts, i.e., a+b=1    ……..(i)

Product of intercepts, i,e., ab=-6   …….(ii)

b=1-a  From(i)

Put b in equation (ii)

a(1-a)=-6

a-a2=-6

a2-a-6=0

a2-3a+2a-6=0

a(a-3)+2(a-3)=0

(a-3)(a+2)=0

a=3 or a=-2

When a=3 in [a+b=1] equation

b=-2

Required equation is xa+yb=1

x3+y-2=1

2x-3y-6=0

When a=-2 in [a+b=1] equation

b=3

Required equation is xa+yb=1

x-2+y3=1

3x-2y+6=0

Q4. Prove that the product of the drawn from the points a2–b2    , 0 and –a2–b2    , 0 to the line xacos +ybsin =1 is b2. [Class 11 Mathematics Chapter 10 Important Questions]

Ans. Given the equation of the line is xacos +ybsin =1

Let p1 be the distance from a2-b2    , 0 to the given line,

p1=a2-b2    a cos – 1cos a2+sin b2  [∵ from the points a2-b2    , 0]

Similarly,

p2 be the distance from -a2-b2    , 0 to the given line,

p2= – a2-b2    a cos – 1cos a2+sin b2

Product of perpendicular lines, i.e., p1p2

p1p2=a2-b2    a cos – 1- a2-b2    a cos – 1cos² a² + sin² b²

=a2-b2a2 cos – 1b2cos² +a2sin² a²b²

= a² cos² – b2cos² -a² a²b²  a²a2sin² +b2cos²

= -a2sin² +b2cos² b2a2sin² +b2cos²   [∵a2cos² -a2=a2cos² -1

= (a2sin² +b2cos² )b2a2sin² +b2cos²

=b2

Hence proved that the product of the drawn from the points a2-b2    , 0 and -a2-b2    , 0 to the line xacos +ybsin =1 is b2.