# Class 11 NCERT Solutions- Chapter 10 Straight Lines – Miscellaneous Exercise on Chapter 10 | Set 1

**Question 1. Find the values of k for which the line (k â€“ 3)x â€“ (4 â€“ k**^{2})y + k^{2} â€“ 7k + 6 = 0 is

^{2})y + k

^{2}â€“ 7k + 6 = 0 is

**(a) Parallel to the x-axis**

**(b) Parallel to the y-axis**

**(c) Passing through the origin.**

**Solution:**

We are given the line, (k â€“ 3)x â€“ (4 â€“ k

^{2})y + k^{2}â€“ 7k + 6 = 0=> (4 â€“ k

^{2})y = (k â€“ 3)x + k^{2}â€“ 7k + 6=> y =+

Now the equation of the line is of the form y = mx + c where m is the slope of the line and c is its y-intercept.

So, we get m =and c =

**(a) Parallel to the x-axis**

If the line is parallel to the x-axis, then we have,

Slope of the line = Slope of the x-axis

=> m = 0

=>= 0

=> k â€“ 3 = 0

=> k = 3

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

**(b) Parallel to the y-axis**

If the line is parallel to the y-axis, then we have,

Slope of the line = Slope of the y-axis

=> m = âˆž (undefined)

=>= âˆž

=> k

^{2}âˆ’ 4 = 0=> k

^{2}= 4=> k = Â±2

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

**(c) Passing through the origin.**

If the line is passing through the origin,

Y-intercept = 0

=> c = 0

=> k

^{2}â€“ 7k + 6 = 0=> (k â€“ 6) (k â€“ 1) = 0

=> k = 1 or k = 6

Therefore, if the given line is passing through the origin, then the value of k is either 1 or 6.

**Question 2. Find the values of Î¸ and p, if the equation x cos Î¸ + y sin Î¸ = p is the normal form of the line âˆš3x + y + 2 = 0.**

**Solution:**

We are given the line, âˆš3x + y + 2 = 0.

=> âˆ’âˆš3x âˆ’ y = 2

On converting this into normal form, we get

=>

=>

=>

By comparing this equation with the given normal form x cos Î¸ + y sin Î¸ = p, we get

=> cos Î¸ =, sin Î¸ =and p = 1

The values of sin Î¸ and cos Î¸ are negative. So, Î¸ = Ï€ +=.

Therefore, the value of Î¸ isand p is 1.

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

**Solution:**

Let’s suppose the intercepts cut by the given lines on the axes are a and b. According to the question, we have,

=> a + b = 1 . . . . (1)

=> ab = â€“ 6 . . . . (2)

By solving both the equations we get

a = 3 and b = â€“2 or a = â€“2 and b = 3

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

bx + ay â€“ ab = 0

When a = 3 and b = â€“2

So the equation of the line is â€“ 2x + 3y + 6 = 0, i.e. 2x â€“ 3y = 6.

When a = â€“2 and b = 3

So the equation of the line is 3x â€“ 2y + 6 = 0, i.e. â€“3x + 2y = 6.

Therefore, the required equation of the lines are 2x â€“ 3y = 6 and â€“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:**

Suppose (0, b) is the point on the y-axis whose distance from line x/3 + y/4 = 1 is 4 units.

The line can be written as 4x + 3y â€“ 12 = 0

By comparing our equation to the general equation of line Ax + By + C = 0, we get

A = 4, B = 3 and C = â€“12

Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from (x

_{1}, y_{1}) is given by,d =

We are given d = 4. For the point (0, b), the value of d becomes,

=>= 4

=>= 4

=> |3b â€“ 12| = 20

=> 3b â€“ 12 = 20 or 3b â€“ 12 = â€“20

=> b = 32/3 or b = â€“8/3

Therefore, (0, 32/3) and (0, â€“8/3) are the points on the y-axis whose distance from the line x/3 + y/4 = 1 is 4 units.

**Question 5. Find the perpendicular distance from the origin to the line joining the points (cos Î¸, sin Î¸) and (cos Ã˜, sin Ã˜).**

**Solution:**

The equation of the line joining the points (cos Î¸, sin Î¸) and (cos Ã˜, sin Ã˜) is given by,

=> y â€“ sin Î¸ =(x â€“ cosÎ¸)

=> y(cos Ã˜ â€“ cos Î¸) â€“ sin Î¸(cos Ã˜ â€“ cos Î¸) = x(sin Ã˜ â€“ sin Î¸) â€“ cos Î¸(sin Ã˜ â€“ sin Î¸)

=> x(sin Ã˜ â€“ sin Î¸) + y(cos Ã˜ â€“ cos Î¸) + cos Î¸ sin Ã˜ â€“ sin Î¸ cos Î¸ â€“ sin Î¸ cos Ã˜ + sin Î¸ cos Î¸ = 0

=> x(sin Ã˜ â€“ sin Î¸) + y(cos Ã˜ â€“ cos Î¸) + sin (Ã˜ â€“ Î¸) = 0

So, we get, A = sin Ã˜ â€“ sin Î¸, B = cos Ã˜ â€“ cos Î¸ and C = sin (Ã˜ â€“ Î¸).

Now, we know that the perpendicular distance(d) of a line Ax + By + C = 0 from the origin (0, 0) is given by,

d =

=

=

=

=

=

=

Therefore,is the perpendicular distance from the origin to the given line.

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

**Solution:**

Two given lines are

x â€“ 7y + 5 = 0 . . . . (1)

3x + y = 0 . . . . (2)

By solving equations (1) and (2) we get

x = âˆ’5/22 and y = 15/22

(âˆ’5/ 22, 15/22) is the point of intersection of lines (2) and (3)

Now the equation of any line parallel to the y-axis is of the form

x = a . . . . (1)

If the line x = a passes through point (âˆ’5/22, 15/22) we get a = âˆ’5/22.

Therefore, the required equation of the line is x = âˆ’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 line is, x/4 + y/6 = 1.

=> 3x + 2y â€“ 12 = 0

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

Here the slope of the given line = âˆ’3/2

So the slope of line perpendicular to the given line = âˆ’1/(âˆ’3/2) = 2/3

Suppose the given line intersects the y-axis at (0, y). So, the equation of the given line becomes,

=> y/6 = 1

=> y = 6

Hence, the given line intersects the y-axis at (0, 6). We know that the equation of the line that has a slope of 2/3 and passes through point (0, 6) is given by,

=> (y â€“ 6) = 2/3 (x â€“ 0)

=> 3y â€“ 18 = 2x

=> 2x â€“ 3y + 18 = 0

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

**Question 8. Find the area of the triangle formed by the lines y â€“ x = 0, x + y = 0 and x â€“ k = 0.**

**Solution:**

It is given that

y â€“ x = 0 . . . . (1)

x + y = 0 . . . . (2)

x â€“ k = 0 . . . . (3)

Here the point of intersection of lines (1) and (2) is x = 0 and y = 0.

And the point of intersection of lines (2) and (3) is x = k and y = â€“ k, lines (3) and (1) is x = k and y = k.

So the vertices of the triangle formed by the three given lines are (0, 0), (k, â€“k) and (k, k).

Here the area of triangle whose vertices are (x

_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) isA =

=

=

=

= k

^{2}square units

Therefore, k^{2}sq. units is the area of the triangle formed by the given lines.

**Question 9. Find the value of p so that the three lines 3x + y â€“ 2 = 0, px + 2y â€“ 3 = 0 and 2x â€“ y â€“ 3 = 0 may intersect at one point.**

**Solution:**

It is given that

3x + y â€“ 2 = 0 . . . . (1)

px + 2y â€“ 3 = 0 . . . . (2)

2x â€“ y â€“ 3 = 0 . . . . (3)

By solving equations (1) and (3) we get

x = 1 and y = â€“1

It is given that the three lines intersect at one point and the point of intersection of lines (1) and (3) will also satisfy line (2).

=> p (1) + 2 (â€“1) â€“ 3 = 0

=> p â€“ 2 â€“ 3 = 0

=> p = 5

Therefore, the value of p is 5.

**Question 10. If three lines whose equations are y = m**_{1}x + c_{1}, y = m_{2}x + c_{2} and y = m_{3}x + c_{3} are concurrent, then show that m_{1} (c_{2} â€“ c_{3}) + m_{2} (c_{3} â€“ c_{1}) + m_{3} (c_{1} â€“ c_{2}) = 0.

_{1}x + c

_{1}, y = m

_{2}x + c

_{2}and y = m

_{3}x + c

_{3}are concurrent, then show that m

_{1}(c

_{2}â€“ c

_{3}) + m

_{2}(c

_{3}â€“ c

_{1}) + m

_{3}(c

_{1}â€“ c

_{2}) = 0.

**Solution:**

It is given that

y = m

_{1}x + c_{1}. . . . (1)y = m

_{2}x + c_{2}. . . . (2)y = m

_{3}x + c_{3}. . . . (3)By subtracting equation (1) from (2), we get,

=> 0 = (m

_{2}â€“ m_{1}) x + (c_{2}â€“ c_{1})=> (m

_{1}â€“ m_{2}) x = c_{2}â€“ c_{1}=> x =

And y =+ c

_{1}=> y =

=> y =

Hence, (,) is the point of intersection of lines (1) and (2).

As the given three lines are concurrent, this point must satisfy the equation (3).

=>

=>

=>

=> m

_{1}(c_{2}â€“ c_{3}) + m_{2}(c_{3}â€“ c_{1}) + m_{3}(c_{1}â€“ c_{2}) = 0

Hence proved.

**Question 11. Find the equation of the lines through the point (3, 2) which **makes** an angle of 45Â° with the line x â€“ 2y = 3.**

**Solution:**

Suppose a is the slope of the line which passes through the point (3, 2).

Given line is x â€“ 2y = 3.

y = 1/2 x â€“ 3/2 which is of the form y = mx + c.

So, the slope of the given line b = 1/2

We know that the angle between the required line and line x â€“ 2y = 3 is 45

^{o}. The angle is given by,tan Î¸ =

=> tan 45

^{0}==>= 1

=>

=> 2 + a = 1 â€“ 2a or 2 + a = â€“ 1 + 2a

=> a = â€“1/3 or a = 3

When a = 3, the equation of the line passing through (3, 2) and having a slope 3 is,

=> y â€“ 2 = 3 (x â€“ 3)

=> y â€“ 2 = 3x â€“ 9

=> 3x â€“ y = 7

When a = â€“1/3, the equation of the line passing through (3, 2) and having a slope â€“1/3 is

=> y â€“ 2 = â€“1/3 (x â€“ 3)

=> 3y â€“ 6 = â€“ x + 3

=> x + 3y = 9

Therefore, the equations of the lines are 3x â€“ y = 7 and x + 3y = 9.

**Question 12. Find the equation of the line passing through the point of intersection of the lines 4x + 7y â€“ 3 = 0 and 2x â€“ 3y + 1 = 0 that has equal intercepts on the axes.**

**Solution:**

Suppose the equation of the line having equal intercepts on the axes as

=> x/a + y/a = 1

=> x + y = a . . . . (1)

By solving equations 4x + 7y â€“ 3 = 0 and 2x â€“ 3y + 1 = 0, we get,

x = 1/13 and y = 5/13

(1/13, 5/13) is the point of intersection of two given lines.

Putting in (1), we get,

=> a = 1/13 + 5/13

=> a = 6/13

Here the equation (1) becomes

=> x + y = 6/13

=> 13x + 13y = 6

Therefore, the required equation of the line is 13x + 13y = 6.

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