Equation Of A Line Concise Solutions Chapter-14 Class 10

Equation Of A Line Concise Solutions Chapter-14 Class 10. Solutions of Exercise – 14 (A), Exercise – 14 (B),Exercise – 14 (C), Exercise – 14 (D),Exercise – 14 (E) of Concise. Concise Selina Maths of ICSE Board Class 10th Solutions for Equation Of A Line Concise Solutions Chapter-14. Step by Step Solutions of Concise Equation Of A LineChapter-14 for ICSE Maths Class 10 is available here. All Solutions of Concise Selina of Chapter-14 Equation Of A Line has been solved according instruction given by council. This is the Solutions of Chapter-14 Equation Of A Line for ICSE Class 10th. ICSE Maths text book of Concise is In series of famous ICSE writer in maths publications. Concise is most famous among students.

Equation Of A Line Concise Solutions Chapter-14 Class 10

The Solutions of Concise Mathematics Equation Of A Line Concise Solutions Chapter-14 for ICSE Class 10 have been solved. Experience teachers Solved Equation Of A Line Concise Solutions Chapter-14 to help students of class 10th ICSE board. Therefore the ICSE Class 10th Maths Solutions of Concise Selina Publishers helpful on various topics which are prescribed in most ICSE Maths textbook

–:Select Topic :–

Exe-14(A), Exe-14(B), Exe-14(C), Exe-14(D), Exe-14(E)

How to Solve Concise Maths Selina Publications Chapter-14 Equation Of A Line

Note:- Before viewing Solutions of Chapter -14 Equation Of A Line of Concise Selina Maths. Read the Chapter Carefully then solve all example of your text book. The Chapter- 14 Equation Of A Line is main Chapter in ICSE board.

Concise Solutions of Exercise -14(A),Equation of a line for ICSE Class 10th Maths

Question 1.

Find, which of the following points lie on the line x – 2y + 5 = 0 :
(i) (1, 3)
(ii) (0, 5)
(iii) (-5, 0)
(iv) (5, 5)
(v) (2, -1.5)
(vi) (-2, -1.5)

Answer 1

Equation of given line x – 2y + 5 = 0

(i)

Substituting x = 1, y = 3, in the given equation.
1 – 2 x 3 + 5 = 0 ⇒ 1 – 6 + 5 = 0 ⇒ 0 = 0, which is true.
(1, 3) satisfies the equation.
(ii)

Substituting x = 0 , y = 5 in the given equation
0 – 2 x 5 + 5 = 0 ⇒ 0 – 10 + 5 = 0 ⇒ -5 = 0, which is not true.
( 0, 5) does not satisfy the equation.

(iii)

Substituting x = – 5, y = 0 in the given equation
-5 – 2 x 0 + 5 = 0 ⇒ -5 – 0 + 5 = 0 ⇒ 0 = 0 which is true.
(-5, 0) satisfies the equation.

(iv)

Substituting x = 5, y = 5 in the given equation.
– 5 – 2 x 5 + 5 = 0 ⇒ -5 – 10 + 5 = 0 ⇒ 0 = 0 which is true.
(5, 5) satisfies the equation.

(v)

Substituting x = 2, y = -1.5 in the given equation.
2 – 2 x (- 1.5) + 5 = 0 ⇒ 2 + 3 + 5 = 0 ⇒ 10 = 0. which is not true.
(2, -1.5) does not satisfy the equation.

(vi)

Substituting x = -2, y = -1.5 in the given equation
– 2 – 2 x (-1.5) + 5 = 0 ⇒ – 2 + 3 + 5 = 0 ⇒ 6 = 0, which is not true.
(-2, -1.5) does not satisfies the equation.

Question 2

State, true or false :
(i) the line $\frac { x }{ 2 }$ + $\frac { y }{ 3 }$ = 0 passes through the point (2, 3).
(ii) the line $\frac { x }{ 2 }$ + $\frac { y }{ 3 }$ = 0 passes through the point (4, -6).
(iii) the point (8, 7) lies on the line y – 7 = 0
(iv) the point (-3, 0) lies on the line x + 3 = 0
(v) if the point (2, a) lies on the line 2x – y = 3, then a = 5.
Solution:

Answer 2

(i)

Equation of the line is $\frac { x }{ 2 }$ + $\frac { y }{ 3 }$ = 0
and co-ordinates of point are (2, 3)
If the point is on the line, then it will satisfy the equation.

(2, 3) is not on the line

(ii)

Equation of the line is $\frac { x }{ 2 }$ + $\frac { y }{ 3 }$ = 0
and co-ordinates of point are (4, -6)
If the point is on the line, then it will satisfy the equation

Hence, point (4, -6) is on the line.

(iii)

Equation of line is y – 7 = 0 and the co-ordinates of point are (8, 7)
If the point is on the line, then it will satisfy the equation
L.H.S. = y – 7 = 7 – 7 = 0 = R.H.S.
Hence, point (8, 7) is on the line.

(iv)

Equation of the line is x + 3 = 0 and co-ordinates of point are (-3, 0)
If the point is on the line, then it will satisfy the equation.
L.H.S. = x + 3 = -3 + 3 = 0 = R.H.S.
Hence, the point (-3, 0) is on the line.

(v)

Equation of the line is 2x – y = 3
and co-ordinates of the point are (2, a)
If the point is on the line, then it will satisfy the equation.
L.H.S. = 2x – y = 2 x 2 – a = 4 – a
R.H.S. = 3
4 – a = 3 ⇒ 4 + 3 = a ⇒ a = 7
But a = 5 given, therefore it is not on the line.
(i) False

(ii) True

(iii) True

(iv) True

(v) False.

Question 3

The line given by the equation 2x – $\frac { y }{ 3 }$ = 7 passes through the point (k, 6); calculate the value of k.

Answer 3

Question 4

For what value of k will the point (3, -k) lie on the line 9x + 4y = 3 ?

Answer 4

Point (3, -k) satisfies the equation 9x + 4y = 3
Substituting x = 3 , y = -k, we get :
9 x 3 + 4 (- k), = 3
⇒ 27 – 4k = 3
and ⇒ – 4k = 3 – 27
so ⇒ – 4k = – 24
hence ⇒ k = 6

Question 5

The line $\frac { 3x }{ 5 }$ – $\frac { 2y }{ 3 }$ + 1 = 0, contains the point (m, 2m – 1); calculate the value of m.

Answer 5

Point (m, 2m -1) satisfies the equation

Question 6

Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2) ?

Answer 6

Line 3x – 5y = 6 bisect the join of points (5, -2) and (-1, 2)
The mid-point of (5, -2) and (-1, 2) satisfies the equation.
Now, mid-point of (5, -2) and (-1, 2)

Now, substituting x = 2, y = 0, in the given equation

Question 7

(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of k.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and ( 0, k). Find the value of k.

Answer 7

(i) line y = 3x – 2 bisects the join of (a, 3) and (2, -5)
Mid-point join of there points satisfies it.
Now, mid-point of (a, 3) and (2, -5) is

Question 8

(i) The point (-3, -2) lies on the line ax + 3y + 6 = 0, calculate the value of ‘a’
(ii) The line y = mx + 8 contains the point (- 4, 4), calculate the value of ‘m’

Answer 8

(i) Point (-3, 2) lies on the line ax + 3y + 6 = 0,
Then x = – 3, y = 2 satisfies it
a (-3) + 3(2) + 6 = 0
⇒ -3a + 6 + 6 = 0
so ⇒ -3a + 12 = 0
therefore ⇒ -3a = – 12
hence ⇒ a = 4
(ii) line y = mx + 8 contains the point (-4, 4)
x = – 4, y = 4 satisfies it
4 = m (-4) + 8
⇒ 4 = -4m + 8
and ⇒ 4m = 8 – 4 = 4
⇒ m = 1

Question 9

The point P divides the join of (2, 1) and (-3, 6) in the ratio 2 : 3. Does P lie on the line x – 5y + 15 = 0 ?

Answer 9

P divides the line joining of the points (2, 1) and (-3, 6) in the ratio of 2 : 3,
co-ordinates of P will be

Now, substituting x = 0, y = 3 in the equation
x – 5y + 15 = 0
⇒ 0 – 5 x 3 + 15 = 0
and ⇒ 0 – 15 + 15 = 0
⇒ 0 = 0 which is true.
Point (0, 3) lies on the line.

Question 10

The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio of 1 : 2. Does the line x – 2y = 0 contain Q ?

Answer 10

Point Q, divides the line segment joining the points (5, -4) and (2, 2) in the rates of 1 : 2
co-ordinates of Q will be,

Now, substituting x = 4, y = – 2 in the equation
x – 2y = 0, we get
4 – 2 x (-2) = 0
⇒ 4 + 4 = 0
⇒ 8 = 0 which is not true.
Point Q does not lie on the line x – 2y = 0

Question 11

Find the point of intersection of the lines : 4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1) x – 2y = 4; find the value of k.

Answer 11

4x + 3y = 1 …..(i)
3x – y = -9 …..(ii).
Multiplying (i) by 1 and (ii) 3
4x + 3y = 1
9x – 3y = -27
Adding, we get-:
1 3x = – 26 ⇒ x = -2
from (ii),
3x – y = – 9
3(-2) – y = -9
⇒ – 6 – y = -9
and ⇒ -y = -9 + 6 = -3
then ⇒ y = 3
The point of intersection is (-2, 3)
The line (2k – 1) x – 2y = 4 passes through that point also
It is satisfy it.
(2k – 1) (-2) – 2(3) = 4
⇒ -4k + 2 – 6 = 4
and ⇒ -4k – 4 = 4
so ⇒ -4k = 4 + 4 = 8
therefore ⇒ k = -2
Hence point of intersection is (-2, 3) and value of k = -2

Question 12

Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.

Answer 12

2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent
They will pass through the same point
Now 2x + 5y = 1 …..(i)
x – 3y = 6 ……(ii)
Multiply (i) by 3 and (ii) by 5, we get :
-6x + 15y = 3
5x – 15y = 30
Adding we get :
11x = 33 ⇒ x = 3
from (ii),
x – 3y = 6
⇒ 3 – 3y = 6
and ⇒ -3y = 6 – 3 = 3
hence ⇒ y = -1
Point of intersection of first two lines is (3, -1)
Substituting the values in third line x + 5y + 2 = 0
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 3 – 5 + 2 = 5 – 5 = 0 = R.H.S.
Hence the given three lines are concurrent.

Concise Solutions of EXERCISE -14 (B), Equation Of A Straight Line Selina Publishers Maths

Question 1

Find the slope of the line whose inclination is :
(i) 0°
(ii) 30°
(iii) 72° 30′
(iv) 46°

Answer 1

(i) Slope of line whose inclination is 0° = tanθ = tan 0° = 0
(ii) Slope of line whose inclination is 30° = tan 30° = $\frac { 1 }{ \surd 3 }$
(iii) Slope of line whose inclination is 72° 30′ = tan 72°30’ = 3. 1716 (Using tables)
(iv) Slope of line whose inclination is 46° = tan 46° = 1.0355 (Using tables)

Question 2

Find the inclination of the line whose slope is:
(i) 0
(ii) √3
(iii) 0.7646
(iv) 1.0875

Answer 2

Slope of a line = tanθ. Where θ is the inclination
(i) When slope is θ. then tanθ = 0 ⇒ θ = 0°
(ii) When slope is θ, then tanθ = √3 ⇒ θ = 60°
(iii) When slope is 0.7646, then tanθ = 0.7646 ⇒ θ = 37°24′ (Using tables)
(iv) When slope is 1.0875, then tanθ = 1.0875 ⇒ θ = 47°24′ (Using tables)

Question 3

Find the slope of the line passing through the following pairs of points :
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b ) and (b, -a)

Answer 3

We know that, slope of a line which passes

Question 4

Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)

Answer 4

Question 5

Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)

Answer 5

Question 6

The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.

Answer 6

Slope of the line passing through two points (0, 2) and (-3, -1)

Question 7

The line passing through (-4, -2) and (2, -3) is perpendicular

Answer 7

Slope of the line passing through the points

Question 8

Without using the distance formula, show that the point A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.

Answer 8

AB and CA are perpendicular to each other
Hence, ΔABC is a right-angled triangle.

Question 9 Equation Of A Line Concise Solutions

Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.

Answer 9

Slopes of AB and DC are equal
AB || DC Similarly slope of BC and slope of DA are equal.
BC || DA
Hence ABCD is a parallelogram.

Question 10

(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the – quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.

Answer 10

QR || PS.
Hence PQRS is a parallelogram.

Question 11 Equation Of A Line Concise Solutions

Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are co llinear.

Answer 11

The given points are P (a, b + c), Q (b, c + a) and R (c, a + b)
We know that, these points P, Q, R are collinear if Slope of PQ = Slope of QR

Slope of PQ = Slope of QR
P, Q and R are collinear.

Question 12

Find x, if the slope of the line joining (x, 2) and (8, -11) is $\frac { -3 }{ 4 }$ .

Answer 12

Slope of line joining (x, 2) and (8, -11) is

Question 13

The side AB of an equilateral triangle ABC is parallel to the x-axis. Find the slopes of all its sides.

Answer 13 Equation Of A Line Concise Solutions

ΔABC is an equilateral

Each angle is equal to 60°
Side AB is parallel to x-axis
Slope of AB = slope of x-axis = 0.
and Slope of AC = tan A = tan 60° = √3
therefore Slope of CB = tan B = tan 120° = tan (180°- 60°) = – tan 60° = -√3
hence Slopes of AB, BC and CA are 0, -√3, √3

Question 14

The side AB of a square ABCD is parallel to the x-axis. Find the slopes of all its sides. Also, find :
(i) the slope of the diagonal AC
(ii) the slope of the diagonal BD.

Answer 14

ABCD is a square in which AB || DC || x-axis.
AD || BC || y-axis
Slope of AB and DC = 0
and slope of AD and BC = not defined (tan90° is not defined)
AC and BD are the diagonals of square ABCD.
Now slope of AC = tan 45° = 1
and slope of BD = tan 135° = tan (180° – 45°) = – tan 45° = -1

Question 15

A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find :
(i) the slope of the altitude of AB
(ii) slope of the median AD and
(iii) the slope of the line parallel to AC.

Answer 15

Vertices of ΔABC are A (5, 4), B (-3, -2), and C (1, -8)

Question 16

The slope of the side BC of a rectangle ABCD is $\frac { 2 }{ 3 }$ . Find
(i) The slope of the side AB,
(ii) the slope of the side AD.

Answer 16

ABCD is a rectangle in which

Question 17 Equation Of A Line Concise Solutions

Find the slope and the inclination of the line AB if
(i) A = (-3, -2) and B = (1, 2)
(ii) A = (0, -√3) and B = (3, 0)
(iii) A = (-1, 2√3) and B = (-2, √3)

Answer 17

Question 18

The points (-3, 2), (2, -1) and (a, 4) are collinear, Find ‘a’.

Answer 18

Points are collinear.
Slope of (-3, 2) and (2, -1) = Slope of (2, -1) and (a, 4)
Now, Slope of (-3, 2) and (2, -1) will be

Question 19

The points (k, 3), (2, -4) and (-k + 1, -2) are collinear. Find k.
Solution:

Answer 19

Points (k, 3), (2, -4) and (-k + 1, -2) are collinear
Slope of (k, 3) and (2, -4) = slope of (2, -4) and (-k + 1, -2)
Now, slope of (k, 3) and (2, -4)
M

Question 20 Equation Of A Line Concise Solutions

Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC ?
Justify your conclusion by calculating the slopes of AB and AC.

Answer 20

Question 21

Find the value(s) of k so that PQ will be parallel to RS. Given :

(i) P (2, 4), Q (3, 6), R (8, 1) and S (10, k)
(ii) P (3, -1), Q (7, 11), R (-1, -1) and S (1, k)
(iii) P (5, -1), Q (6, 11), R (6, -4k) and S (7, k²)

Answer 21