Respuesta :
Answer:
The equation of the line is y + x + 1 = 0.
Step-by-step explanation:
The given equations are : x + 2y = 0 and 3x + y + 5 = 0
Now, finding the intersection point of the above system:
from (1) , x = -2y
put in (2), 3 (-2y) + y + 5 = 0
or, 5y = 5 ,or y = 1
If y = 1, pitting in (1), x = -2
So, the intersection lines is (-2,1).
the other point on line is (-3,2)
Now, finding the slope m of the line : [tex]m =\frac{y_2 -y_1}{x_2 - x_1} =\frac{2 -1}{-3 - (- 2)} [/tex]
or, [tex]m = -\frac{1}{1}[/tex] = -1
So, by POINT SLOPE FORM: the equation of a line is
(y - y0) =m (x -x0),
now for (-3,2) : equation is ( y - 2) = (-1) (x +3)
or, y + x + 1 = 0
Hence, the equation of the line that passes through (−3, 2) and the intersection of the lines x+2y=0 and 3x+y+5=0 is y + x + 1 = 0.
Answer:
Hence the Equation of line with points ( - 3 , 2) and slop - 1 is Y + X + 1 = 0 Answer
Step-by-step explanation:
Given two line equation as
x + 2y = 0 And
3x + y + 5 = 0
Now, intersection point of this two lines is :
Solve the given two line equations
Multiply eq 1 by 3,
So ,
3x + 6y = 0
3x + y = - 5
Again , (3x + 6y) - (3x + y) = 5
Or, 5y = 5
I.e y = 1
Put the value of y in above eq
So, 3x + ( 1) = - 5
Or, 3x + 1 = -5
I.e 3x = - 6
So, x = -2
Now equation of line with points ( - 3, 2 ) and ( -2 , 1) is
First we find the slop ( m ) = [tex]\frac{(y2 - y1)}{(x2 - x1)}[/tex]
m = [tex]\frac{(1 - 2)}{(- 2 + 3)}[/tex]
Or, m = - 1
Or, m = -1
∴ Equation of line with points ( - 3 , 2) and slop -1 is
Y - y1 = m (X - x1)
Y - 2 = -1 (X + 3)
Or, Equation of line is Y + X + 1 = 0
Hence the Equation of line with points ( - 3 , 2) and slop - 1 is Y + X + 1 = 0 Answer
