KEAM 2010 Math Paper

The slope of line 3x−y=−5 is m1=3
∴ The slope of required line is
=‌

m1±tan‌45∘

1∓m1⋅tan‌45∘

=‌

3±1

1∓3⋅1

=−2,

1

2

Alternate
Given line 3x−y=−5
m1=3
Let slope of another line =m2
Now, angle between both the lines
tan‌θ=|‌

m1−m2

1+m1m2

|
tan‌45∘=‌

m1−m2

1+m1m2

or ‌

m2−m1

1+m1m2

(If m1>m2)
⇒‌1+m1m2=m1−m2
⇒‌1+3m2=3−m2
⇒‌m2=‌

1

2

(If m2>m1)
Then, 1+3m2=m2−3
⇒‌‌m2=−2