Question

Find the equations of the straight lines passing through the point of intersection of x+y+1=0, x-y+1=0 and whose distance from (-1, 6) is 2.​

Answer 1

Answer:

Equation of all lines passing through intersection of given lines is L: 3x-4y+4+k(x+y-1) =0 for different values of k. It's slope is (3+k)/(-4+k). Since L is perpendicular to x-y=0, whose slope is 1. So

(3+k)/(-4+k) *1 = -1 or or 3+k=4-k => 2k=1 or k=1/2

So reqd line is 3x-4y+4+(1/2)*(x+y-1)=0. Simplifing

7x-7y+7=0 or x-y+1=0

Answer 2

Equation of all lines passing through intersection of given lines is L: 3x-4y+4+k(x+y-1) =0 for different values of k. It's slope is (3+k)/(-4+k). Since L is perpendicular to x-y=0, whose slope is 1. So

(3+k)/(-4+k) *1 = -1 or or 3+k=4-k => 2k=1 or k=1/2

So reqd line is 3x-4y+4+(1/2)*(x+y-1)=0. Simplifing

7x-7y+7=0 or x-y+1=0

Answer 3

Equation of all lines passing through intersection of given lines is L: 3x-4y+4+k(x+y-1) =0 for different values of k. It's slope is (3+k)/(-4+k). Since L is perpendicular to x-y=0, whose slope is 1. So

(3+k)/(-4+k) *1 = -1 or or 3+k=4-k => 2k=1 or k=1/2

So reqd line is 3x-4y+4+(1/2)*(x+y-1)=0. Simplifing

7x-7y+7=0 or x-y+1=0