When the equation of a line is1) Find the equation of a straight line that passes through the point (2,3) and is parallel to the line with equation 3x + 4y -8 = 0.

ax+by +c = 0

where a,b and c can be positive or negative

its slope is given by(-b)/a

-for being parralel slopes should be same

-for two lines being perpendicular the slopes will multiply to give-1

it can also be represented as

y= -bx/a +c

any point(m,n)lying on this should satisfy the equation

which means whenever you will put

am+bn+c

the value you will get is"0"

1)

Let the line be given by

y = mx+c

Since it is parallel to

3x+4y-8 =0

its slope will be same

Thus the value of "m" is-4/3

y= (-4x/3) + c

Put the value of (x,y) as (2,3)

Thus

3 = -8/3 + c

Thus c= 17/3

y=-4x/3 + 17/3

2)

y= mx +c

since it is perpendicular to

4x + 3y = 6.

Thus

m * -3/4 = -1

Hence

m= 4/3

Thus

y= (4x/3) +c

4 = -8/3 +c

c= 20/3

y= 4x/3 +20/3