KCET 2017 Math Solved Paper

Given line,y=3x−1 →(1)
General equation of line is given by
y=mx+c
Here m = 3 and c = –1.
Condition for two perpendicular lines is
m1⋅m2=−1
So, 3m2=−1⇒m2=−‌

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Standard equation of line passing through point (x1,y1) is given by (y−y1)=m(x−x1)
Here (x1,y1)=(1,2).So,
(y−2)=−‌

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(x−1)
⇒−3y+6=x−1
⇒x+3y−7=0