CBSE Important Questions Class 11 Maths Chapter 9 – Straight Lines

Class 11 Maths Chapter 9 Straight Lines is part of the CBSE Class 11 Maths syllabus. It explains the slope of a line, the conditions for two lines to be parallel or perpendicular, and the angle between two lines. Students learn the different forms of the equation of a line — point-slope, two-point, slope-intercept, intercept and normal form — and how to find the distance of a point from a line.

These CBSE important questions for Class 11 Maths Chapter 9 cover finding the slope from two points, writing the equation of a line in various forms, the angle between two lines, the condition for perpendicular lines, the distance of a point from a line, and the distance between two parallel lines. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Find the slope of the line passing through the points (3, −2) and (−1, 4).

Answer: The slope of a line through (x₁, y₁) and (x₂, y₂) is
m = (y₂ − y₁) / (x₂ − x₁)
= (4 − (−2)) / (−1 − 3)
= 6 / (−4)
= −3/2
Since the slope is negative, the line slopes downwards from left to right.

Q2. Write the different forms of the equation of a straight line.

Answer: • Point-slope form: y − y₁ = m(x − x₁)
• Two-point form: y − y₁ = [(y₂ − y₁)/(x₂ − x₁)](x − x₁)
• Slope-intercept form: y = mx + c, where c is the y-intercept
• Intercept form: x/a + y/b = 1, where a and b are the x and y intercepts
• Normal form: x cosω + y sinω = p, where p is the perpendicular distance from the origin
• General form: Ax + By + C = 0

Q3. Find the equation of the line passing through (−2, 3) with slope −4.

Answer: Using the point-slope form y − y₁ = m(x − x₁):
y − 3 = −4(x − (−2))
y − 3 = −4(x + 2)
y − 3 = −4x − 8
4x + y + 5 = 0

Q4. Find the equation of the line passing through the points (1, −1) and (3, 5).

Answer: First find the slope:
m = (5 − (−1)) / (3 − 1) = 6/2 = 3
Now use the point-slope form with the point (1, −1):
y − (−1) = 3(x − 1)
y + 1 = 3x − 3
3x − y − 4 = 0

Q5. Write the condition for two lines to be (i) parallel and (ii) perpendicular.

Answer: Let the slopes of the two lines be m₁ and m₂.
(i) Parallel: The lines are parallel when their slopes are equal, that is m₁ = m₂.
(ii) Perpendicular: The lines are perpendicular when the product of their slopes is −1, that is m₁m₂ = −1.
In this case each slope is the negative reciprocal of the other.

Q6. Find the angle between the lines y − √3x − 5 = 0 and √3y − x + 6 = 0.

Answer: Writing both in slope-intercept form:
First line: y = √3x + 5, so m₁ = √3
Second line: y = x/√3 − 2, so m₂ = 1/√3
Using tanθ = |(m₁ − m₂)/(1 + m₁m₂)|:
= |(√3 − 1/√3) / (1 + 1)|
= |(3 − 1)/√3 ÷ 2|
= (2/√3)/2 = 1/√3
So θ = 30°

Q7. Find the distance of the point (3, −5) from the line 3x − 4y − 26 = 0.

Answer: The distance of a point (x₁, y₁) from the line Ax + By + C = 0 is
d = |Ax₁ + By₁ + C| / √(A² + B²)
Here A = 3, B = −4, C = −26 and the point is (3, −5).
d = |3(3) − 4(−5) − 26| / √(9 + 16)
= |9 + 20 − 26| / 5
= 3/5
= 0.6 units

Q8. Find the distance between the parallel lines 3x − 4y + 7 = 0 and 3x − 4y + 5 = 0.

Answer: For two parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0:
d = |C₁ − C₂| / √(A² + B²)
Here C₁ = 7, C₂ = 5, A = 3 and B = −4.
d = |7 − 5| / √(9 + 16)
= 2/5
= 0.4 units

Q9. Reduce the equation 3x − 4y + 10 = 0 into slope-intercept form and find its slope and y-intercept.

Answer: Rearrange to make y the subject:
3x − 4y + 10 = 0
−4y = −3x − 10
y = (3/4)x + 10/4
y = (3/4)x + 5/2
Comparing with y = mx + c:
Slope m = 3/4 and y-intercept c = 5/2

Q10. Find the equation of the line that cuts off intercepts 3 and −2 on the x-axis and y-axis respectively.

Answer: Using the intercept form x/a + y/b = 1, with a = 3 and b = −2:
x/3 + y/(−2) = 1
x/3 − y/2 = 1
Multiplying throughout by 6:
2x − 3y = 6
2x − 3y − 6 = 0

Q11. Find the equation of the line passing through (−3, 5) and perpendicular to the line 2x − 3y + 7 = 0.

Answer: The slope of the given line: 3y = 2x + 7, so y = (2/3)x + 7/3, giving m₁ = 2/3.
For a perpendicular line, m₁m₂ = −1, so m₂ = −3/2.
Using the point-slope form with the point (−3, 5):
y − 5 = (−3/2)(x + 3)
2(y − 5) = −3(x + 3)
2y − 10 = −3x − 9
3x + 2y − 1 = 0

Q12. Find the slope of a line which makes an angle of 30° with the positive direction of the x-axis. Also write the slope of a line parallel to the y-axis.

Answer: The slope of a line is m = tanθ, where θ is the angle it makes with the positive x-axis.
For θ = 30°:
m = tan 30° = 1/√3
For a line parallel to the y-axis: Such a line makes an angle of 90° with the x-axis, and tan 90° is not defined.
So the slope of a vertical line is undefined. (A line parallel to the x-axis has slope 0.)

Related Study Material on Extramarks

Q.1 The sum of first 24 terms of the A.P. a1, a2, a3,, if it is given that a1+ a5+ a10+ a15+ a20+ a24 = 225, is

Marks:1

Ans 900.

We know that in an A.P. the sum of the terms equidistance from the beginning and end is always same and is equal to the sum of first and last term i.e.,

a1+ an = a2+ an-1= a3+ an-2 =

So, if an A.P. consists of 24 terms, then

a1+ a24= a5+ a20= a10+ a15.

Now, a1+ a5+ a10+ a15+ a20+ a24 = 225

i.e.,(a1+ a24)+(a5+a15)+( a10+ a15) = 225

3(a1+ a24) = 225

a1+ a24 = 75 (1)

Therefore, S24 = (24/2) (a1+ a24)

= 900.

Q.2 The fourth term of a G.P. is 6, then the product of its first eight terms is

Marks:1

Ans

Let a be the first term and r the common ratio.

Therefore, ar3= 6 (1)

Now, product of first eight terms

a(ar)(ar2)(ar3)(ar4) (ar5) (ar6) = a7r21

= (ar3)7

= 67.

Q.3 Find the sum of the series:0.7 + 0.77 + 0.777 + to n terms

Marks:6

Ans

We have 0.7+0.77+0.777+ to n terms=710+77100+7771000+ to n terms=7110+11100+1111000+ to n terms On multiply and divide by 9 we get, =79910+99100+9991000+ to n terms=7910110+1001100+100011000+ to n terms=791110+11100+111000+ to n terms=791+1+1+ to n terms110+1100+11000+ to n terms=79n110+1102+1103+ to n terms Here, we get GP with a=110 and r=110Sn=a1rn1r=79n1101110n1110=79n110—1091110n1=79n191110n

Q.4 If a, b, c are in G.P., then prove that a2 b2, b2 c2,  c2 a2 are in G.P.

Marks:4

Ans

a, b, c are in G.P. Let r be the common ratio. Then
b = ar and c = ar2
a2 b2, b2 c2, c2 a2 will be in G.P. if

b2-c2a2-b2=c2-a2b2-c2

Substituting b = ar and c = ar2 we get

ar2ar22a2ar2 =ar22ar32ar2ar22a2r21r2a21r2 =a2r41r2a2r21r2r2 =r2

Hence a2 b2, b2 c2, c2 a2 is a G.P.

Q.5 A snail travels in a straight line path. Assume that it travels 1cm in the first second and half of the distance traveled in the previous second in the succeeding second. In how many seconds would it reach a point 31/16 cm away Thus calculate the speed of the snail.

Marks:6

Ans

Distance traveled by snail in the first second = 1 cm
Distance traveled in the second sec.

=1—12=12cm

Distance traveled in the third sec

=12—12=14cm

Thus the distances covered form a G.P. 1, 1/2, 1/4,…
Here, a = 1 and r

=12.

Let n seconds be the required time. Therefore,

Sum =1112n112=311612n=132n=5secnow, speed=distance· time=3116·5=3180cm/sec

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