Question 523223: what is the standard form of the equation of the line passing through the point (-2,-2) and perpindicular to the line 3x-5y=-5 ? Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website! what is the standard form of the equation of the line passing through the point (-2,-2) and perpindicular to the line 3x-5y=-5 ?
Standard Form of Equation of the line:
y=mx+b
Given
3x-5y=-5
rearrage the above equation according to the standard form
-5y=-3x-5
-5y/-5=-(3x+5)/-5
y=3/5(x)+1
Compare above equation with the standard form equation
m=3/5 and b=1
Since lines are perpendicular multiplicatin of their slope will be (-1)
So slope of the required line will be (-5/3)
Now we have a point(-2,-2) and slope (-5/3)of the line we can easily find required lines by putting these values in the equation of the straight line poin-slope form.
m=(y2-y1)/(x2-x1)
-5/3=(y-(-2))/(x-(-2))
-5/3=(y+2))/(x+2)
-5(x+2)=3(y+2)
-5x-10=3y+6
-3y=5x+10+6
-3y=5x+16
-3y/-3 = (5x+16)/-3
y=-5/3(x)-16/3
Above equation is the required equation of the line in standard Form.
Red line = Given line
Green line = Required line
