1.

Find the equation of straight line (1) parallel to 3x + 4y +5=0(2) perpendicular to 3x+4y+5=0and passing through the point (2,-1)​

Answer»

Step-by-step explanation:

3x+4y+5=0. (1)

it PASSES through (2,-1)and has slop m1

if it is parallel to new line than

m1=m2

so new line

3x+4y+c=0

3(2)+4(-1)+c=0

6-4+c=0

2+c=0

c=-2

the line parallel to

3x+4y+5=0 and PASSING through (2,-1) is

3x+4y+c=0

put the value of c

3x+4y+(-2)=0

3x+4y-2=0

(2.) perpendicular to 3x+4y+5=0

y=mx+c

3x+5=-4y

y=(-3/4)X+(-5/4) (2)

if these are perpendicular m1*m2=-1

(-3/4)*m2=-1

m2=4/3

perpendicular line will be

y=(4/3)x-5/4=0

or

12y=16x-15

or

16x-12y-15=0

Now passing through (2,-1)

16x-12y-c

32+12=c

c=48

eq 16x-12y-48

I think it is helpful for you



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