find equation of line parallel to line 5x+3y+1=0 and having y intercept 8
Answers
Answer:
Hi hope this helps....
Step-by-step explanation:
given equation: 5x+3y+1=0
equation parallel to this can be written as: 5x+3y+a=0
given required equation y intercept is 8
now converting it into intercept form we have x/-a/5 +y/-a/3=1
now equating y intercept with 8
-a/3=8
-a=24
a=-24
therefore, required eq 5x+3y-24=0
Answer: The given line
The given line5 x - 3 y + 1 = 0
The given line5 x - 3 y + 1 = 0can be rewritten as
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1or
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) is
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomes
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomesy - (-3) = (- 3/5) (x -4)
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomesy - (-3) = (- 3/5) (x -4)=> y + 3 = -3 x/5 + 12/5
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomesy - (-3) = (- 3/5) (x -4)=> y + 3 = -3 x/5 + 12/5=> 5 y + 15 = - 3 x + 12
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomesy - (-3) = (- 3/5) (x -4)=> y + 3 = -3 x/5 + 12/5=> 5 y + 15 = - 3 x + 12=> 3 x + 5 y + 3 = 0
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomesy - (-3) = (- 3/5) (x -4)=> y + 3 = -3 x/5 + 12/5=> 5 y + 15 = - 3 x + 12=> 3 x + 5 y + 3 = 0The required equation is:
The given line5 x - 3 y + 1 = 0can be rewritten as3 y = 5 x + 1ory = (5/3) x + ⅓ —————————————-(1)has a slope = 5/3The slope of the straight line perpendicular to the given line = - 3/5The equation of straight line with slope m and passing through the point (x¹, y¹) isy - y¹ = m (x - x¹).In our case (x¹, y¹) = (4, - 3) and the equation of the required line becomesy - (-3) = (- 3/5) (x -4)=> y + 3 = -3 x/5 + 12/5=> 5 y + 15 = - 3 x + 12=> 3 x + 5 y + 3 = 0The required equation is:3 x + 5 y + 3 = 0.