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).

Answer 1

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

Answer 2

The slope of the line $x+4y=6$ is $m=-\frac{1}{4}$.

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

$y=-\frac{1}{4} x+c$

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

$5=-\frac{1}{4} (-8)+c \\ c=3$

Thus the equation of the line in slope intercept form is $y=-\frac{1}{4} x+3$

The slope of the line $2 x-3y=12$ is $m=\frac{2}{3}$.

The line perpendicular to the above line has slope $-\frac{1}{m} =-\frac{3}{2}$

Its equation is $y=-\frac{3}{2} x+c$. Since this line passes through, $(2,6)$,

$6=-\frac{3}{2} (2)+c\\c=9$.

Thus the equation of the line in slope intercept form is $y=-\frac{3}{2} x+9$