Equation of a Line Exe-14A Class-10 Concise ICSE Maths Solution Ch-14. In this article you would learn basic concept of equation of a line. We Provide Step by Step Solutions / Answer of questions for Selina Concise Maths. Visit official Website CISCE for detail information about ICSE Board Class-10 Mathematics.
Equation of a Line Exe-14A Class-10 Concise ICSE Maths Solution Ch-14
| Board | ICSE |
| Publications | Selina |
| Subject | Maths |
| Class | 10th |
| Chapter-14 | Equation of a Line |
| Writer | R.K. Bansal |
| Exe-14A | basic concept of equation of a line |
| Edition | 2025-2026 |
Basic Concept of Equation of a Line
Equation of a Line Exe-14A Class-10 Concise ICSE Maths Solution Ch-14.
Que-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)
Ans: 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.
Que-2: State, true or false :
(i) the line x/2 + y/3 = 0 passes through the point (2, 3).
(ii) the line x/2 + 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:
Ans: (i) False.
Explanation: The given line is x2+y3=0
Substituting x = 2 and y = 3 in the given equation,
L.H.S = x2+y3
= 1 + 1
= 2 ≠ R.H.S
(ii) True.
Explanation: The given line is x2+y3=0
Substituting x = 4 and y = – 6 in the given equation,
L.H.S = 42+-63
= 2 − 2
(iii) True
Explanation: 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) True
Explanation: 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) False
Explanation: 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.
Que-3: The line given by the equation 2x – y/3= 7 passes through the point (k, 6); calculate the value of k.
Ans: Given, the line given by the equation 2x-y3=7 passes through the point (k, 6).
Substituting x = k and y = 6 in the given equation, we have:
2x-6/3=7
2k – 2 = 7
2k = 9
k = 9/2
k = 4.5
Que-4: For what value of k will the point (3, -k) lie on the line 9x + 4y = 3 ?
Ans: 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
Que-5: The line 3x/5 – 2y/3 + 1 = 0, contains the point (m, 2m – 1); calculate the value of m.
Ans: The equation of the given line is 3x/5-2y/3+1=0
Putting x = m, y = 2m − 1, we have:
3m/5-2(2m-1)/3+1=0
3m/5-4m-2/3+1=0
9m-20m+10/15=-1
9m – 20m + 10 = –15
–11m = –25
m = 25/11
Que-6: Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2) ?
Ans: Line 3x – 5y = 6 bisect the join of points (5, -2) and (-1, 2), if the co-ordinates of the mid-point of AB satisfy the equation of the line.
The co-ordinates of the mid-point of AB are
(5-1/2,-2+2/2)
= (4/2,0/2)
= (2, 0)
Substituting x = 2 and y = 0 in the given equation, we have:
L.H.S. = 3x − 5y
= 3(2) − 5(0)
= 6 − 0
= 6 = R.H.S.
Hence, the line 3x − 5y = 6 bisect the join of (5, −2) and (−1, 2)
Que-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.
Ans: The given line bisects the join of A(a, 3) and B(2, −5), so the co-ordinates of the mid-point of AB will satisfy the equation of the line.
The co-ordinates of the mid-point of AB are
(a+2/2,3-5/2)
= (a+2/2,-2/2)
= (a+2/2,-1)
Substituting x=a+2/2 and y = –1 in the given equation, we have:
y = 3x – 2
-1=3×a+2/2-2
3×a+2/2=1
a+2=2/3
a=2/3-2
= 2-6/3 = -4/3
Que-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’
Ans: (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
Que-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 ?
Ans: P divides the line joining of the points (2, 1) and (-3, 6) in the ratio of 2 : 3,
Co-ordinates of the point P are
(2×(-3)+3×22+3,2×6+3×12+3)
= (-6+65,12+35)
= (05,155)
= (0, 3)
Substituting x = 0 and y = 3 in the given equation, we have:
L.H.S. = 0 − 5(3) + 15
= −15 + 15
= 0 = R.H.S.
Hence, the point P lies on the line x − 5y + 15 = 0.
Que-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 ?
Ans: Given, Point Q, divides the line segment joining the points (5, -4) and (2, 2) in the rates of 1 : 2
Co-ordinates of the point Q are
(1×2+2×5/1+2 , 1×2+2×(-4)/1+2)
= (2+10/3,2-8/3)
= (12/3,-6/3)
= (4, –2)
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
Que-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.
Ans: 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
Que-12: Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Ans: 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.
–: Equation of a Line Exe-14A Class-10 Concise ICSE Maths Solution Ch-14 :–
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