**NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3**

NCERT Solutions for Class
11 Maths Chapter 10 Straight
Lines Ex 10.3 are the part of NCERT Solutions for Class 11 Maths. Here you can
find the NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3.

**Ex 10.3
Class 11 Maths Question 1.**

Reduce the following equations into slope - intercept form and find their
slopes and the y-intercepts.**(i)**x + 7y = 0,

**(ii)**6x + 3y – 5 = 0,

**(iii)**y = 0

**Solution:**

**(i)** The given equation is x + 7y = 0, which can be written in the
form

7y = –x ⇒ y = −x/7 + 0 … (1)

Also, the slope - intercept form is y = mx + c …(2)

On comparing (1) and (2), we get

m = −1/7, c = 0

Hence, the slope is −1/7 and the y-intercept = 0.

**(ii)** The given equation is 6x + 3y – 5 = 0, which
can be written in the form

3y = –6x + 5

⇒ y = –2x + 5/3 …(1)

Also, the slope - intercept form is y = mx + c … (2)

On comparing (1) and (2), we get

m = –2 and c = 5/3

i.e. slope = –2 and the y-intercept = 5/3

**(iii)** The given equation is y = 0, which
can be written in the form

y = 0·x + 0 … (1)

Also, the slope - intercept form is y = mx + c … (2)

On comparing (1) and (2), we
get

m = 0, c = 0.

Hence, slope is 0 and the y-intercept is 0.

**Ex 10.3 Class 11 Maths Question 2.**

Reduce the following equations into intercept form and find their intercepts on
the axes.**(i)**3x + 2y – 12 = 0,

**(ii)**4x – 3y = 6,

**(iii)**3y + 2 = 0

**Solution:**

**(i)** The
given equation is 3x + 2y – 12 = 0

We have to reduce the given equation into intercept
form,

i.e., x/a + y/b = 1 …(1)

Now, the given equation can be written as 3x + 2y = 12

⇒ 3x/12 + 2y/12 = 1 ⇒ x/4 + y/6 = 1 …(2)

On comparing (1) and (2), we get a = 4, b = 6

Hence, the intercepts of the line are 4 and 6.

**(ii)** The given equation is 4x – 3y = 6

We have to reduce the given equation into intercept form,

i.e., x/a + y/b = 1 …(1)

Now, the given equation can be written as

4x/6 − 3y/6 = 1 or 2x/3 – y/2 = 1 …(2)

On comparing (1) and (2), we get

a = 3/2 and b = –2

Hence, the intercepts of the line are 3/2 and –2.

**Ex 10.3 Class 11 Maths Question
3.**

Reduce the
following equations into normal form. Find their perpendicular distances from
the origin and angle between perpendicular and the positive x-axis.**(i)**x – √3y + 8 = 0,

**(ii)**y – 2 = 0,

**(iii)**x – y = 4.

**Solution:**

**Ex 10.3 Class 11 Maths Question
4.**

Find the
distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2).

**Solution:**

**Ex 10.3 Class 11 Maths Question 5.**

Find the points on the x-axis, whose distances from the line x/3 + y/4 = 1 are 4 units.

**Solution:**

The given equation of line is x/3 + y/4 = 1, which can be
written as

4x + 3y – 12 = 0 … (i)

Let (a, 0) be the point on x-axis whose distance from line (i) is 4 units.

**Ex 10.3 Class 11
Maths Question 6.**

Find the distance between parallel lines**(i)**15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

**(ii)**l(x + y) + p = 0 and l(x + y) – r = 0.

**Solution:**

If the lines are Ax + By + C_{1} = 0 and Ax + By + C_{2} =
0, then distance between

**Ex 10.3 Class 11 Maths Question 7.**

Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing
through the point (-2, 3).

**Solution:**

The given equation of line is 3x – 4y + 2 = 0 … (i)

This equation can be written as y = 3x/4 + ½

Slop of the line = 3/4

Thus, slope of any line parallel to the given line (i) is 3/4 and passes through (-2, 3), then its
equation is

**Ex 10.3 Class 11 Maths Question 8.**

Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x
intercept 3.

**Solution:**

The given equation of line is x – 7y + 5 = 0 … (i)

This equation can be written as y = x/7 + 5/7

Slope of this line = 1/7

Thus, slope of any line perpendicular to the line (i) is -7 and passes through
(3, 0), then its equation is

y – 0 = -7(x – 3) [∵ Product of slope of perpendicular lines
is -1]

⇒ y = -7x + 21

⇒ 7x + y – 21 = 0

This is the required equation of line.

**Ex 10.3 Class 11 Maths Question 9.**

Find angles between the lines √3x + y = 1
and x + √3y = 1.

**Solution:**

The given equations are

√3x + y = 1 … (i)

x + √3y = 1 … (ii)

Since we have to find the angle between the two lines, therefore, firstly we need
to find the slopes of lines (i) and (ii).

**Ex 10.3 Class 11
Maths Question 10.**

The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19
= 0 at right angle. Find the value of h.

**Solution:**

The given points are (h, 3) and (4, 1).

∴ Slope of the line joining the points (h,
3) and (4, 1)

**Ex 10.3 Class 11 Maths Question 11.**

Prove that the line through the point (x_{1}, y

_{1}) and parallel to the line Ax + By + C = 0 is A(x – x

_{1}) + B(y – y

_{1}) = 0.

**Solution:**

The given equation of line is Ax + By + C = 0

∴ Slope of the above line = −A/B

i.e. slope of any line parallel to given line and passing through (x_{1},
y_{1}) is −A/B.

Then, equation is (y – y_{1}) = −A/B (x – x_{1})

⇒ B(y – y_{1}) = -A(x – x_{1})

⇒ A(x – x_{1}) + B(y – y_{1})
= 0

Hence proved.

**Ex 10.3 Class 11
Maths Question 12.**

Two lines passing through the point (2, 3) intersects each other at an angle of
60°. If slope of one line is 2, find equation of the other line.

**Solution:**

We have given that two lines are passing through the point (2, 3) and intersects
each other at an angle of 60°.

Let m be the slope of the other line.

**Ex 10.3 Class 11
Maths Question 13.**

Find the equation of the right bisector of the line segment joining the points
(3, 4) and (-1, 2).

**Solution:**

Let the given points are A(3, 4) and B(-1, 2).

Let M be the mid-point of AB.

**Ex 10.3 Class 11 Maths Question 14.**

Find the coordinates of the foot of perpendicular from the point (-1, 3) to the
line 3x – 4y – 16 = 0.

**Solution:**

The given line is 3x – 4y – 16 = 0
… (i)

Slope of the line (i) = 3/4

Then, equation of any line perpendicular from (-1, 3) to the given line (i) is:

**Ex 10.3 Class 11
Maths Question 15.**

The perpendicular from the origin to the line y = mx + c meets it at the point
(-1, 2). Find the values of m and c.

**Solution:**

We have given that the
perpendicular from the origin to the line y = mx + c meets it at the point (-1,
2).

∴ 2 = m(-1) + c … (i)

⇒ c – m = 2

**Ex 10.3 Class 11
Maths Question 16.**

If p and q are the lengths of perpendiculars from the origin to the lines x cos
Î¸ – y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k, respectively, prove that p^{2}+ 4q

^{2}= k

^{2}.

**Solution:**

We have given that p and q are the lengths of perpendiculars from the origin to
the lines x cos Î¸ – y sin Î¸ = k cos 2Î¸ and x sec Î¸ + y cosec Î¸ = k.

**Ex 10.3 Class 11
Maths Question 17.**

In the triangle ABC with vertices A(2, 3), B(4, -1) and C( 1, 2), find the
equation and length of altitude from the vertex A.

**Solution:**

We have given a AABC with the vertices, A (2, 3), B (4, -1) and C (1, 2)

**Ex 10.3 Class 11 Maths Question 18.**

If p is the length of perpendicular from the origin to the line whose
intercepts on the axes are a and b, then show that 1/p^{2}= 1/a

^{2}+ 1/b

^{2}.

**Solution:**

We have given that p is the length of perpendicular from the origin to the line
whose intercepts

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