KCET 2018 Math Solved Paper

Given line,3x−4y+2=0 → (1)
⇒y=‌

3

4

x+‌

1

2

So, m=

3

4

and c=

1

2

Since line is parallel then slope of both lines should same. Then, equation of parallel is given by (y−y1)=‌

3

4

(x−x1)
And this line passes through point (−2, 3). Then,
(y−3)=‌

3

4

(x−(−2))
⇒4y+12=3x+6
⇒3x−4y+18=0
Therefore, required equation of line is 3x−4y+18=0