Verified answer

1. y = -2x + 4

2. y = 2/5x - 1/5

(a) Given line is y=12-2x that is 2x+y-12=0

Any line parallel to ax+by+c=0 is of the form ax+by+k=0

So required line is of the form 2x+y+k=0

But it passes through (-1,6)

So substitute the point in the above line equation 2(-1)+6+k=0 => k= - 4

So the required line equation is 2x+y-4=0

(b) Given line is 2x+5y=14 that is 2x+5y-14=0

Any line perpendicular to ax+by+c=0 is of the form bx-ay+k=0

So required line is of the form 5x-2y+k=0

But it passes through (2,-1)

So substitute the point in the above line equation 5(2)-2(-1)+k=0 => k= -12

So the required line equation is 5x-2y-12=0

Thats it..

You can do any sum in the similar way..

Thank you..

(a) Equation of any line parallel to the line

ax+by+c=0

and passing through the point (p, q) is given by

a(x-p)+b(y-q)=0

The given line is

y=12-2x

This eqn can rewritten as

2x+y-12=0

The required line is parallel to the above line and passes through the point (-1, 6). Hence its equation is

2(x-(-1))+1(y-6)=0

i.e., 2x+y-4=0

Answer(b):

Any line perpendicular to the line

ax+by+c=0

and passing through the point (p, q) is given by

b(x-p)-a(y-q)=0

The required line is perpendicular to the line

2y+5x=14

i.e., 5x+2y-14=0

and passes through (2, -1).

Hence, its equation is

2(x-2)-5(y-(-1))=0

i.e., 2x-5y-9=0

Hope that helps