Question

Find the equation of a straight line passing through the point (-1,3) and parallel to x+y+1=0 through slope point form.

Answer 1

given:

point:(-1,3)

equation:x+y+1=0

to find:

equation of the line using slope point form.

formula:

༒slope point form:y-y1=M(x-x1)

༒slope(M)=-a/b

solve:

slope of the equation x+y+1=0

slope(M)=-1/1

=-1

now we have slope and point(-1,3)

slope point form:

y-3=-1(x-[-1])

y-3=-1(x+1)

y-3=-x-1

take the -x-1 left side

x+y-2=0

result:

the equation of the line is x+y-2=0.

I hope it is useful for you.plz mark in brain list

Answer 2

Solution:-

As we know, slope of the line ax+by+c=0, is -a/b

so,the slope of linex+y+1=-1

since the line is parallel,to this line,,so

it will have the same slope

considering the slope,

let the eqaution of required line be x+y+c=0

since the line is passing through (-1,3) ,so putting this vaalue of x and y we get

-1+3+c=0

c=-2

so,the equation of line will be x+y-2=0

{hope it helps}