Respuesta :
Let us find the slope of the line : x+ 4y = 6
we can write it as 4y = -x +6
y = -(1/4) x + ( 6/4)
slope of the line y= mx + b is m
so here slope = m = -1/4
because we have to find a paraller line so the slope for required line would be same : -1/4
equation of the line having slope m and passing through x1 y1 is :
Y= m( X-x1) + y1
let's plug m= -1/4 x1= -8 and y1 = 5
Y= (-1/4) ( X- -8) + 5
Y= ( -1/4) ( x+8) + 5
Y= (-1/4)( X ) + (-1/4)( 8) + 5
Y= (-1/4) x + 3
Let us now work on second part :
given line is : 2x- 3y = 12
let's write it -3y = -2x +12
y= (-2/-3) x + ( 12/ -3)
Y= ( 2/3) x - 4
slope of this line is 2/3
the required line is perpendicular to it
so the slope of required line = negative resiprocal of this slope =
(-1/ slope of this line ) = -1/( 2/3) = -3/2
Equation of line with slope m= -3/2 and passing through x1= 2 and y1 = 6 is :
Y= m( X-x1) + y1
Y= ( -3/2) ( X- 2) + 6
Y= (-3/2) X - 2(-3/2) + 6
Y= ( -3/2 ) X + 9
Answer : for part 1 : Y= ( 2/3) x - 4
Answer for part 2 : Y= ( -3/2 ) X + 9
The slope of the line [tex] x+4y=6 [/tex] is [tex] m=-\frac{1}{4} [/tex].
The equation of the line in slope intercept form parallel to [tex] x+4y=6 [/tex] is
[tex] y=-\frac{1}{4} x+c [/tex]
Since the above line passes through [tex] (-8,5) [/tex] is
[tex] 5=-\frac{1}{4} (-8)+c \\ c=3 [/tex]
Thus the equation of the line in slope intercept form is [tex] y=-\frac{1}{4} x+3 [/tex]
The slope of the line [tex]2 x-3y=12 [/tex] is [tex] m=\frac{2}{3} [/tex].
The line perpendicular to the above line has slope [tex] -\frac{1}{m} =-\frac{3}{2} [/tex]
Its equation is [tex] y=-\frac{3}{2} x+c [/tex]. Since this line passes through, [tex] (2,6) [/tex],
[tex] 6=-\frac{3}{2} (2)+c\\c=9 [/tex].
Thus the equation of the line in slope intercept form is [tex] y=-\frac{3}{2} x+9 [/tex]