Question

On a graph paper draw two straight lines which real monenl X-3y + 1 = 0 and 2x - 3y - 4=0. Also find the point of intersection of the two lines the graph.​

Answer 1

Answer:

Equation of family of lines through two lines L

1

and L

2

are L

1

+λL

2

=0

L

1

=>2x+3y−4=0

L

2

=>x−5y−7=0

L

1

+λL

2

=0

=>2x+3y−4+λ(x−5y−7)=0

=>x(2+λ)+y(3−5λ)−4−7λ=0

x−intercept =

2+λ

4+7λ

Given x−intercept =−4

∴

2+λ

4+7λ

=−4

=>λ=

11

−12

Therefore, equation of line is 2x+3y−4−

11

12

(x−5y−7)=0

=>10x+93y+40=0