obtain the equation of two points passing of lines 4x-3y-1=0 and 2x-3y+3=0 and equally inclined to the axes
1 answer:
Answer:By 8(4)P.5, the slope of the line equally inclined to axes is tan(±45
o
)=±1. Hence by P+λQ=0,
5λ+3
2λ+4
=±1
⇒λ=
3
1
,−1. Putting for λ, the two lines are
x−y=0,x+y−2=0.
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