Question

Find the equation of the line that passes through (−3, 2) and the intersection of the lines x+2y=0 and 3x+y+5=0.

Answer 1

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 ) = $\frac{(y2 - y1)}{(x2 - x1)}$

                                     m = $\frac{(1 - 2)}{(- 2 + 3)}$

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

Answer 2

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 : $m =\frac{y_2 -y_1}{x_2 - x_1} =\frac{2 -1}{-3 - (- 2)}$

or, $m = -\frac{1}{1}$ = -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.