NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3 are part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines Ex 10.3.
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 11 |
| Subject | Maths |
| Chapter | Chapter 10 |
| Chapter Name | Straight Lines |
| Exercise | Ex 10.3 |
| Number of Questions Solved | 18 |
| Category | NCERT Solutions |
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) We have given an equation x + 7y = 0, which can be written in the form
⇒ 7y = – x ⇒ y = \(\frac { -1 }{ 7 } \)x + 0 … (1)
Also, the slope intercept form is y-mx + c …(2)
On comparing (1) and (2), we get
m = \(\frac { -1 }{ 7 } \), c = 0
Hence the slope is \(\frac { -1 }{ 7 } \) and the y-intercept = 0.
(ii) We have given an equation 6x + 3y – 5 = 0, which can be written in the form 3y = – 6x + 5
⇒ y = – 2x + \(\frac { 5 }{ 3 } \) …(1)
Also, the slope intercept form is y = mx + c … (2)
On comparing (1) and (2), we get
m = – 2 and c = \(\frac { 5 }{ 3 } \)
i.e. slope = – 2 and the y-intercept = \(\frac { 5 }{ 3 } \)
(iii) We have given an equation y = 0
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) Given equation is 3x + 2y – 12 = 0 We have to reduce the given equation into intercept form, i.e., \(\frac { x }{ a } +\frac { y }{ b } =1\) …(1)
Now given, 3x + 2y = 12
⇒ \(\frac { 3x }{ 12 } +\frac { 2y }{ 12 } =1\quad \) ⇒ \( \quad \frac { x }{ 4 } +\frac { y }{ 6 } =1\) …(2)
On comparing (1) and (2), we get a = 4, b = 6 Hence, the intercepts of the line are 4 and 6.
(ii) Given equation is 4x – 3y = 6
We have to reduce the given equation into intercept form, i.e., \(\frac { x }{ a } +\frac { y }{ b } =1\) …(1)
\(\frac { 4 }{ 6 } x-\frac { 3 }{ 6 } y=1\quad or\quad \frac { x }{ 3/2 } +\frac { y }{ -2 } =1\) …(2)
On comparing (1) and (2), we get
a = \(\frac { 3 }{ 2 }\) and b = – 2
Hence, the intercepts of the line are \(\frac { 3 }{ 2 }\) and -2.
(iii) Given equation is 3y + 2 = 0
We have to reduce the given equation into intercept form, i.e., \(\frac { x }{ a } +\frac { y }{ b } =1\)
3y = -2
⇒ y = \(\frac { -2 }{ 3 } \)
The above equation shows that, it is not the required equation of the intercept form as it is parallel to x-axis.
We observe that y-intercept of the line is \(\frac { -2 }{ 3 } \), but there is no intercept on x-axis.
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 – \(\sqrt { 3 } y\) + 8 = 0,
(ii) y-2 = 0,
(iii) x-y = 4.
Solution:
(i) Given equation is x – \(\sqrt { 3 } y\) + 8 = 0
\(\sqrt { 3 } y\) x – \(\sqrt { 3 } y\) = -8
\(\sqrt { 3 } y\) -x + \(\sqrt { 3 } y\) = 8 … (i)
Also, \(\sqrt { \left( coeff.\quad of\quad { x }^{ 2 } \right) +\left( coeff.of\quad { y }^{ 2 } \right) } \)
\(\sqrt { \left( 1 \right) ^{ 2 }+\left( \sqrt { 3 } \right) ^{ 2 } } \quad =\quad \sqrt { 1+3 } =\sqrt { 4 } =2\)
Now dividing both the sides of (1) by 2, we get
\(-\frac { 1 }{ 2 } x+\frac { \sqrt { 3 } }{ 2 } y=4\)
⇒ – cos 60°x + sin 60° y = 4.
⇒ {cos (180° – 60°)) x + {sin (180° – 60°)|y = 4
⇒ cos 120° x + sin 120° y = 4
∴ x cos 120° + y sin 120° = 4 is the required equation in normal form
∵ The normal form is x coso) + y sin⍵ = p
So, ⍵ = 120° and p = 4
⍵ Distance of the line from origin is 4 and the angle between perpendicular and positive x-axis is 120°.
(ii) Given equation is y – 2 = 0
⇒ y = 2
⇒ 0 · x + l · y = 2
⇒ x cos 90° + y sin 90° = 2 is the required equation in normal form
∵ The normal form is x cos⍵ + y sin⍵ = p
So, ⍵ = 90° and p = 2
⍵ Distance of the line from origin is 2 and the angle between perpendicular and positive x-axis is 90°.
(iii) Given equation is x – y = 4 … (1)
Also \(\sqrt { \left( coeff.\quad of\quad { x }^{ 2 } \right) +\left( coeff.of\quad { y }^{ 2 } \right) } \)
\(\sqrt { \left( 1 \right) ^{ 2 }+\left( -1 \right) ^{ 2 } } =\sqrt { 1+1 } =\sqrt { 2 } \)
Now dividing both the sides of (1) bt \(\sqrt { 2 } \), we get
is the required equation in normal form.
∵ The normal form is x cos⍵ + y sin⍵ = p
So, P = \(2\sqrt { 2 } \) and ⍵ = 315°
∴ Distance of the line from the origin is \(2\sqrt { 2 } \) and the angle between perpendicular and the positive x-axis is 315°.
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:
The equation of line is 12(x + 6) = 5(y – 2) …(i)
⇒ 12x + 72 = 5y-10
⇒ 12x – 5y + 82 = 0
∴ Distance of the point (-1, 1) from the line (i)
\(=\frac { \left| 12\left( -1 \right) -5\left( 1 \right) +82 \right| }{ \sqrt { \left( 12 \right) ^{ 2 }+\left( -5 \right) ^{ 2 } } } =\frac { 65 }{ 13 } =5units\)
Ex 10.3 Class 11 Maths Question 5.
Find the points on the x-axis, whose distances from the line \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\) are 4 units.
Solution:
We have a equation of line \(\frac { x }{ 3 } +\frac { y }{ 4 } =1\), 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 = 0and 15x + 8y+31 =0
(ii) |(x + y) + p = 0 and |(x + y) – r = 0.
Solution:
If lines are Ax + By + Q = 0
and Ax + By + C2 = 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:
We have given an equation of line 3x – 4y + 2 = 0
Slop of the line(i) = \(\frac { 3 }{ 4 } \)
Thus, slope of any line parallel to the given line (i) is \(\frac { 3 }{ 4 } \) 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:
Given equation is x – 7y + 5 = 0 … (i)
Slope of this line = \(\frac { 1 }{ 7 } \)
∴ Slope of any line perpendicular to the line (i) is -7 and passes through (3, 0) then
y – 0 = -7(x – 3)
[∵ Product of slope of perpendicular lines is -1]
⇒ y = -7x + 21
⇒ 7x + y – 21 = 0, is the required equation of line.
Ex 10.3 Class 11 Maths Question 9.
Find angles between the lines \(\sqrt { 3 } x\) + y = 1 and x + \(\sqrt { 3 } y\) = 1.
Solution:
The given equations are
\(\sqrt { 3 } x\) + y = 1 … (i)
x + \(\sqrt { 3 } y\) = 1 … (ii)
Since we have to find an angle between the two lines i.e., firstly we have to find the slopes of (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:
Given points are (h, 3) and (4,1).
∴ Slope of the line joining (h, 3) & (4,1)
Ex 10.3 Class 11 Maths Question 11.
Prove that the line through the point (x1 y1) and parallel to the line Ax + By + C = 0 is A(x-x1) + B(y-y1) = 0.
Solution:
Given equation of a line is Ax + By + C = 0
∴ Slope of the above line = \(\frac { -A }{ B } \)
i.e. slope of any line parallel to given line and passing through (x1, y1) is \(\frac { -A }{ B } \)
Then equation is (y – y2) = \(\frac { -A }{ B } \) (x – x1)
=> B(y – y1) = -A(x – x1)
=> A(x – x1) + B(y – y1) = 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 a point (2, 3), through which two lines are passing and intersects 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:
suppose the given points are A and B.
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:
We have, 3x – 4y – 16 = 0
Slope of the kine(i) = \(\frac { 3 }{ 4 } \)
Then equation of any line ⊥ 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:
Given, 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 cosθ – y sinθ = k cos 2θ and x secθ + y cosecθ = k, respectively, prove that p2 + 4q2 = k2.
Solution:
Given p and q are the lengths of perpendiculars from the origin to the lines x cos θ – ysinθ=k cos 2θ and xsecθ+y cosec θ = k.
Ex 10.3 Class 11 Maths Question 17.
In the triangle ABC with vertices A(2, 3), 8(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 \(\frac { 1 }{ { p }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } } +\frac { 1 }{ { b }^{ 2 } } \).
Solution:
Given, p be the length of perpendicular from the origin to the line whose intercepts
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