Selina Concise Class 10 Math Chapter 14 Equation of a line Exercise 14A Solutions
Equation of a line
Exercise 14A
(Q.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)
Solution:
i) Given equation of line is x – 2y + 5 = 0
Now, Check whether (1, 3) lie on this given line.
So, put x = 1 and y = 3 in the given equation.
∴ LHS = x – 2y + 5
= 1 – 2 × (3) + 5
= 1 – 6 + 5
= 6 – 6
= + 0
= R.H.S
L.H.S = R.H.S
∴ The points (1, 3) lie on the equation of x – 2y + 5 = 0
(ii) (0, 5)
Check whether the points (0, 5) lie on the given equation of line.
L.H.S = x – 2y + 5
Put x = 0 and y = 5
= 0 = 2 × 5 + 5
= -10 + 5
= – 5
L.H.S ≠ R.H.S
∴ The points (0, 85) is not lie on the given equation of line x – 2y + 5 = 0
(iii) (-5, 0)
Check whether (-5, 0) lie on the given equation of line.
Put x = -5 and y = 0
L.H.S = x – 2y + 5
= – 5- 2 × 0 + 5
= -5 -0 + 5
= 0
∴ L.H.S = R.H.S
∴ The points (-5, 0) lie on the given equation of line.
(iv) (5, 5)
Check whether (5 5) lie on the given equation of line.
Put x = 5 and y = 5
L.H.S = x – 2y + 5
L.H.S = 5 – 2 × 5 + 5
= 5 – 10 + 5
= 10 – 10
= 0
L.H.S = R.H.S
∴ The points (5, 5) lie on the given equation of line.
(v) (2, -1.5)
Check whether (2, -1.5) lie on the given equation of line.
L.H.S = x – 2y + 5
Put x = 2 and y = -1.5
L.H.S = 2 – 2× (-1.5) + 5
= 2 + 3 + 5
= 5 + 5
= 10
L.H.S ≠ R.H.S
∴ The points (2, -1.5) is not lie on the given equation of line.
(vi) (-2, -1.5)
Check whether (-2, -1.5) lie on the given equation of line.
L.H.S = x – 2y + 5
put x = – 2 and y = – 1.5
= – 2 – 2 × ( – 1.5) + 5
= – 2 + 3 + 5
= 3 + 3
= 6
L.H.S ≠ R.H.S
∴ The points (- 2, – 1.5) is not lie on the given line
Q.2) State, true of 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:-
i) given equation of line
x/2 + y/3 = 0
it passes through the point (2, 3).
we have to check the point passes trough (2, 3)
So, put x = 2 and y = 3 in given equation.
L.H.S = x/2 + y/3
= 2/2 + 3/3
= 1 + 1
= 2
L.H.S ≠ R.H.S
∴ The given statement is wrong.
(ii) given equation of line is
x/2 + y/3 = 0
it passes through the point (4 ,- 6)
we have the check the point passes trough (4, – 6)
So, put x = 4 and y = – 6
L.H.S = x/2 + y/3
= 4/2 + (-6/3)
= 2 – 2
= 0
L.H.S = R.H.S
∴ The given statement is true
(iii) given equation of line is y – 7 = 0
we have to check (8, 7) lies on the given equation of line.
So, substitute y = 7 in given equation
L.H.S = 7 – 7
= 0
∴ L.H.S = R.H.S
∴ The given statement is true
(iv) given equation of line is
x + 3 = 0
we have to check (- 3, 0) lies on the line, x + 3 = 0
So, substitute x = – 3
L.H.S = x + 3
= – 3 + 3
= 0
L.H.S = R.H.S
∴ The given statement is true
(v) given equation of line is 2x – y = 3
we have to check (2,a) lies on the given equation of line
2x – y = 3
So, put x = 2 and y = a
∴ 2 (2) – y = 3
4 – a = 3
4 – 3 = a
1 = a
∴ The given statement is False.
because the value of a = 5
Q.3) The line given by the equation 2x – y/3 = 7 passes through the point (k, 6) ; calculate the value of k.
= Solution
given the equation of line is
2x – y/3 = 7
This equation of line passes through the point (k. 6)
we have to find the value of k = ?
So, put x = k and y = 6
in given equation.
2 (k) – y/3 = 7
2k – 6/3 = 7
2k – 2 = 7
2k = 7 + 2
2k = 9
[k = 9/2]
∴ the value of k is 9/2
Q.4) For what value of k will the point (3 ,- k) lie on the line 9x + 4y = 3 = ?
Solution:-
given equation of line is 9x + 4y = 3
given that (3, – k) lie on the line
put x= 3 and y = – k
9 (3) + 4 (- k) = 3
we have to find the value of k = ?
27 – 4k = 3
27 – 3 = 4k
24/4 = k
6 = k
∴ The value of k is 6
Q.5) The line 3x/5 – 2y/3 + 1 = 0
contains the point (m, 2m – 1),
calculate the value of m
Solution:-
The given equation of line is 3x/5 – 2y/3 + 1 = 0
Also given that the points (m, 2m – 1)
So, put x = m and y = 2m – 1
3(m)/5 – 2 (2m-1)/3 + 1 = 0
3m/5 – (4m – 2/3) + 1 = 0
3m/5 – 4m – 2/3 + 1 = 0
3 × 3m – 5 (4m – 2) /15 = – 1
9m – 20m + 10/ 15 = – 1
– 11m + 10 = – 1 × 15
– 11m + 10 = – 15
– 11m = – 15 – 10
– 11m = – 25
[m = 25/11]
Here is your solution of Selina Concise Class 10 Math Chapter 14 Exercise 14A Equation of a line
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