math help please!!?

Question

can u please give me the steps on how to solve these??

1.write an equation of the line perpendicular to 5y=-x+1 passing through (2,5)

2.solve for x: 6x^2+13x-5=0

3.equation of perpendicular bisector of segment AB?

how do u find the negative reciprical?

Answer 1

1.Set = y: y = -1/5x+1/5 which means the slope of the perpendicular is 5. Using y-y1=m(x-x1), y-5=5(x-2)
distribute the 5 and get y-5=5x-10. Add 5 to each side and y=5x-5. That's in slope-intercept form. Standard form would be:5x-y=5

2.factors into: (3x-1)(2x+5)=0 If 3x-1=0 then x = 1/3. If 2x+5=0 then x=-5/2

3. You find the slope of AB then find its negative reciprical. You find the midpoint of AB. Using y-y1=m(x-x1), you substitute the negative reciprocal slope for m and the midpoint for (x1,y1) then simplify like your problem #1.

juliagulia has the formula wrong, that's why her answer is not right.

Good luck

TO FIND THE NEGATIVE RECIPRICAL, YOU REVERSE THE NUMERATOR AND DENOMINATOR AND TAKE THE OPPOSITE SIGN. SO FOR 2/3 IT WOULD BE -3/2. FOR -3 IT WOULD BE 1/3. FOR 1 IT WOULD BE -1. fOR 4/3 IT WOULD BE -3/4.......continued good luck

Answer 2

1. i think for the 1st question u isolate for y so y=x/5 + 1/5
that means the slope is 1/5, prependicular is the reciprocal so 1/5 becomes 5 but then u need to make that negative so slope (m)= -5 then u substutitute it to the equation y +y1= m (x+x1) (the ones after the x and y are a subscript and m is ur slope -5)
sosubstitute in the (2,5) into the equation:

y+ 5= -5(x+2) then just isolate for y
y= -5x-10-5
y= -5x+15

or put it into standard form 0= -5x-y+15 but the first # cant be (-) so 0=5x+y-15

Answer 3

first no 2 : 6x^2-2x+15x-5=0
2x(3x-1)+5(3x-1)=0
(2x+5) (3x-1) = 0
x = -5/2 or 1/3

Now 1 y = x/5 + 1/5 y = mx+c here m = 1/5 slope of line 1/5 So slope of per line will be -5
eqn of per line y-y1 = m(x-x1) here y1=2 m = -5 and x1=5 now u can slove