Question

Write an equation of a line in slope intercept form that is parallel to the line x + 4y = 6 and passes through (-8, 5).

Show work.

Write an equation of a line in slope intercept form that is perpendicular to the line 2x -3y = 12 and passes through the point (2, 6).
Show work.

Answer 1

The slope of the line
`x+4y=6` is
`m=-(1)/(4)`.

The equation of the line in slope intercept form parallel to
`x+4y=6` is

`y=-(1)/(4) x+c`

Since the above line passes through
`(-8,5)` is

`5=-(1)/(4) (-8)+c \\ c=3`

Thus the equation of the line in slope intercept form is
`y=-(1)/(4) x+3`

The slope of the line
`2 x-3y=12` is
`m=(2)/(3)`.

The line perpendicular to the above line has slope
`-(1)/(m) =-(3)/(2)`

Its equation is
`y=-(3)/(2) x+c`. Since this line passes through,
`(2,6)`,

`6=-(3)/(2) (2)+c\\c=9`.

Thus the equation of the line in slope intercept form is
`y=-(3)/(2) x+9`

Answer 2

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