Find the equation, in standard form, of the line perpendicular to x – 5y = 4 and passing through (-3, 5)?

Question

With all integer coefficients.

Answer 1

x-5y=4
adding -x
-5y=-x+4
dividing by -5
y=(1/5)x-(4/5)
slope of the given line=1/5
so slope of perpendicular line=-5
the line passes through(-3,5)
sothe equation is
y-5=-5(x+3)
y-5=-5x-15
5x+y+10=0 is the required equation of the line

Answer 2

Lines perpendicular to x - 5y = 4 have the form 5x + y = c (since if ax + by = c, -a/b is the slope, and perpendicular lines have slopes that are negative reciprocals).

So just substitute (-3,5) into 5x + y = c and get 5(-3) + 5 = c = -10.

So your equation is 5x + y = -10.

Answer 3

The slope of the given line is 1/5 so the slope of the perpendicular is -5. 5 = -5(-3) +b so b = -10.

This makes the slope intercept equation: y = -5x-10 and the standard form would be 5x + y = -10

Answer 4

The slopes of perpendicular lines are negative inverses of each other. To find the slope of this line,
x - 5y = 4
-5y = -x + 4
y = 1/5 x -4/5
The slope of the line perpendicular to this is the slopes negative inverse; the slope of this line is 1/5, so the slope of the line perpendicular to the line is - 5/1 = -5. To find the y intercept of the line perpendicular to it use the following formula: y- y1 = m(x - x1)
y - 5 = -5(x --3)
y - 5 = -5(x +3)
y - 5 = -5x - 15
y = -5x - 10
In standard form,
y + 5x = -10

Answer 5

The equation of a line perpendicular to x-5y=4 is
Let y =mx+b which passes thru (-3,5)
so 5= -3m+b

Here m=(b-5)/3

The slope of firs line is m1 = 1/5
From condition of perpendicular m1*m= -1
so m*1/5=-1
m= -5=(b-5)/3
b= -10
Equation of that line is y= -5x-10
i. e 5x+y+10=0 Ans.

Answer 6

y= -1/5x+6