# Important Questions Class 11 Maths 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.

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.

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.

[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.

Q.1 A straight line through P(3, 5) is such that its intercept

between the axes is bisected at P. Its equation is

Marks:1

Ans5x + 3y  30 = 0.

Q.2

If the lines y = 5x + 1 and y = x + 1 are equally inclined to the line y = mx + 9, then the value of m is

Marks:1

Ans

1

Q.3

Find the equation of the straight line which passes through the origin and making an angle of 60° with the line 3x + 4y = 2.

Marks:6

Ans

Slope of the given line 3x+4y=2 is (3)/4. Let the slope of the required line be m. tan=m2m11+m1m2Here, m1=34 and m2=m and Î¸=60°. Putting these values in the above formula tan60°=m341+34—m=3m+34=313m4and m+34=313m4m+33m4=334and m+33m4=3+34m(4+33)4=4334and m(4+33)443+34m=4334+33and m=43+34+33Using y=mx+c and c=0 (as the line passes through the origin), we get y=4334+33x+0=4334+33x and y=43+34+33x+0=43+34+33x(433)x(4+33)y=0 and (43+3)x(4+33)y=0

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### 1. If the angle between two lines is 𝞹4 and the slope of one of the lines is 12, find the slope of the other line.

We know that the acute angle between two lines with slopes m1 and  m2 is given by

tan =m1m21+m1m2                                                                                    ………..(1)

Let m1=12m2=m, and =𝞹4 .

Now, putting these values in (1), we get

tan𝞹4=m – 121+12m     or  1=m – 121+12m

which gives  m – 121+12m=1 or m – 121+12m=-1.

Therefore, m=3 or   m=-13

Hence, the slope of the other line is 3 or 13 .

### 2. Reduce the equation 3x+y-8=0 into normal form. Find the values of p and .

. Given equation is

3x+y-8=0                                                                      ……….(1)

Dividing (1) by 32   +12

32x + 12y=4 or cos 30°x+sin 30°y=4.               ……….(2)

Comparing (2) with x cos + y sin =p,  we get p=4 and =30°.

### 3. Show that the path of a moving point such that its distances from two lines 3x-2y=5 and 3x+2y=5 are equal in a straight line.

Given lines are

3x-2y=5                                                     ………(1)

and              3x+2y=5                                                      ……..(2)

Let h,k is any point, whose distances from lines (1) and (2) are equal. Therefore

3h-2k-59+4=3h+2k-59+4  or  3h-2k-5=3h+2k-5,

which gives 3h-2k-5=3h+2k-5  or -3h-2k-5=3h+2k-5.

Solving these two relations we get k=0 or h=53. Thus, the point h,k satisfies the   equations y=0 or x=53, which represent straight lines. Hence, the path of the point equidistant from lines (1) and (2) is a straight line.

### 4. Which of the topics in Class 11 Mathematics Chapter 10 Straight Lines are important?

The topics of Class 11 Mathematics Chapter 10 Straight Lines that are important are:

• Two-point form
• Slope-intercept form
• Intercept form
• Normal form
• The slope of a line when the coordinates of any two points on the line are given
• Conditions for parallelism and perpendicularity of lines in terms of their slopes
• The angle between two lines
• Collinearity of three points
• Different forms of Ax + By + C = 0
• Distance of a Point From a Line
• Distance between two parallel lines
• Various Forms of the Equation of a Line
• Horizontal and vertical lines
• Point-slope form
• The slope of the line
• General Equation of a Line