Question

12.
The equation of the line parallel to x + 3y + 1 = 0 and passing through the
point where 3x - 2y + 5 = 0) cuts the x-axis is:
5
A.
x + 3y
0
B.
5x - 3y + - = 0
3.
5
x-3y = 0
3

x-3y + = 4
x-3y = =
5
x + 3y +
3
LC
= 0
D.​

Answer 1

Answer:

12

Step-by-step explanation:

done. 100% correct

Answer 2

Answer:

The equation of required line is 3x+9y+5=0

Step-by-step explanation:

The equation of line parallel to

$x + 3y + 1 = 0$

is

$x + 3y + c = 0$

where c is some constant

Now to find the point where the line 3x-2y+5=0 cuts the x-axis,we have to take y=0

==>3x+5=0

$x = \frac{ - 5}{3}$

also given that the line x+3y+c=0 passes through this above point

therefore,

$\frac{ - 5}{3} + 2(0) + c = 0 \\ = = > c = \frac{5}{3}$

now substitute the value if c in our required equation

$x + 3y + \frac{5}{3} = 0 \\ 3x + 9y + 5 = 0$

hence,our required equation is 3x+9y+5=0