NCERT Solutions For Class 11 Maths chapter-10 Straight Lines Miscellaneous Exercise

NCERT Solutions For Class 11 Maths chapter-10 Straight Lines Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter-10 Straight LInes

Academic team of Physics Wallah developed step by step NCERT Solutions for Class 11 Maths Chapter-10 Straight LInes Miscellaneous Exercise according to recommen dations and Guideline of CBSE. You can download and share NCERT Solutions for Class 11 Maths.

Question 1. Find the values of k for which the line (k−3)x − (4−k ² )y + k ² −7k + 6=0

(a) Parallel to the x-axis,

(b) Parallel to the y-axis,

(c) Passing through the origin.

Solution :
Given: Equation of line (k−3)x − (4−k ² )y + k ² −7k + 6=0

If the line is parallel to x-axis,

Slope of the line = Slope of the x-axis

(a) If the line parallel to x-axis, then lope of the line = Slope of the x-axis

it can be writen as (k−3)x − (4−k ² )y + k ² −7k + 6=0

we get

Which is of the form y = mx + c

Here, the slope of the given line

Consider the slope of x-axis =0
= 0

k−3=0

k=3

Therefore, if the given line is parallel to the x
-axis, then the value of k is 3
.

(b) If the line parallel to y-axis, then

The given equation of line is (k−3)x − (4−k ² )y + k ² −7k + 6=0

Here if the line is parallel to the y-axis, it is vertical and the slope will be undefined.

So, the slope of the given line =

Here, (k−3) / (4−k ² ) is undefined at k2=4

k ² = 4

⇒k=±2

Therefore, if the given line is parallel to the y
-axis, then the value of k is ±2.

(c) If the line passes through origin then

The given equation of line is (k−3)x − (4−k ² )y + k ² − 7k + 6 = 0

Here, if the line is passing through (0,0)

which is the origin satisfies the given equation

of line,

(k−3)(0) − (4−k ² ) (0)+k ² −7k+6=0

k ² −7k+6=0

Separate the terms,

k ² −6k−k+6=0

(k−6)(k−1)=0

k=1or 6

Therefore, if the given line is passing through the origin, then the value of k

is either

1or 6

Question 2. Find the values of θ and p, if the equation x cos θ +  y sinθ = p is the normal form of the line √3x + y + 2=0.

Solution :
Given: √3x + y + 2=0.

3x + y = -2

Dividing both sides by ,

Question 3. Find the equations of the lines which cut-off intercepts on the axes whose sum and product are 1 and -6 respectively.

Solution :Consider, the intercepts cut by the given lines on the axes are a
and b.

a + b = 1→(1)(1)

ab = − 6 → (2)

Solve both the equations to get

a = 3 and b = −2 or a = −2 and b = 3

We know that the equation of the line whose intercepts on a and b axes is

xa  +  yb  =  1

or

bx  + ay − ab = 0

Case 1:

a = 3
and b = −2

Now, the equation of the line is −2x + 3y + 6 = 0

That is, 2x − 3y = 6

Case 2:

a=−2
and b=3

Now, the equation of the line is 3x − 2y + 6 = 0

That is, −3x + 2y = 6

Therefore, the required equation of the lines are  2x − 3y = 6
and −3x + 2y = 6.

Question 4. What are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units?

Solution :
Given: Equation of line (0,b) and let point on x/3 + y/4 = 1 axis be 4

it can be writen  as 4x + 3y -12 = 0 → (1)

compare equation (1) to the perpendicular distance  (d) of a line

4x + 3y -12 = 0 A = 4,B = 3 and C = −12

We know that the perpendicular distance (d) of a line Ax+By+C=0 from (x1,y1) is,

Question 5. Find perpendicular distance from the origin to the line joining the points (cosθ, sin θ) And (cosϕ,sinϕ).

Solution :
Equation of the line joining points (cosθ, sin θ) and (cosϕ,sinϕ)

y(cosϕ − cosθ) − sinθ(cosϕ − cosθ)

=x(sinϕ − sinθ) − cosθ(sinϕ − sinθ)x (sinθ − sinϕ) − y(cosϕ − cosθ) + cosθ sinϕ − cosθsinθ − sinθ cosϕ  + sinθ cosθ

=0x(sinθ − sinϕ) + y(cosϕ − cosθ) + sin(ϕ−θ)=0

Ax + By + C=0, where A

= sinθ − sinϕ,B= cosϕ − cosθ,  and C=sin(ϕ−θ)

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x ₁ , y ₁ ) is given by d= |Ax ₁ + By ₁ + C| / √A ₂ +√B ₂ .

Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is

Question 6. Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

Solution :

The equation of any line parallel to the y-axis is of the form x = a → (1)

The two given lines are x−7y + 5 = 0 → (2)

3x + y = 0 → (3)

Solve equation (2) and (3)
, we get x = −5 / 22 and y = 15 / 22


Thus, (−5 / 22 , 15 /22)
is the point of intersection of lines (2) and (3).

Since line x = a passes through point (−5 / 22,15 / 22),

a = −5 /22

Therefore, the required equation of the line is x = −5 22

Question 7. Find the equation of a line drawn perpendicular to the line x/4 + y/6 = 1 through the point, where it meets the y-axis.

Solution :
Given: Equation of line x/4 + y/6 = 1

Here, the equation of the given line is x / 4 + y / 6 = 1

This equation can be written as 3x + 2y − 12 = 0

Rewrite as,

y=−3 / 2x + 6
, which is of the form y = mx + c

Now, Slope of the given line  = −3 / 2

∴Slope of line perpendicular to the given line = −1 / (−3/2 ) = 2 / 3



Let the given line intersect the y-axis at (0,y)

.

Substitute x with 0 in the equation of the given line,

y / 6 = 1

⇒ y = 6

∴ The given line intersects the y-axis at (0,6)

The equation of the line that has a slope of 23 and passes through point (0,6) is,(y−6)=23(x−0) Cross multiply and expand brackets,

3y − 18 = 2x

2x − 3y + 18 = 0

Therefore, the required equation of the line is 2x − 3y + 18 = 0
.