Question

The line 3x-4y+7=0 is rotated through an angle 45degree in the clockwise direction about the point(-1,1). The equation of the line in its new position is

Answer 1

Given line equation,
3x-4y+7=0
slope of above line = 3/4
Given that, the line is rotated through an angle 45degree.
Therefore angle between original line and the line formed after rotating is 45degree.
Let the slope of rotated line be m.
We know that,
tanA=I[(m1-m2)/(1+m1m2)]I

tan45°=I[(3/4-m)/(1+3m/4)]

1=I[(3-4m)/(4+3m)]I

±1=(3-4m)/(4+3m)
Squaring on both sides we get,
1=(9+16m²-24m)/(16+9m²+24m)

16+9m²+24m=9+16m²-24m

7m²-48m-7=0

7m²-49m+m-7=0

7m(m-7)+1(m-7)=0

(7m+1)(m-7)=0

m=7 and m=-1/7
By slope point form we can form the equations.
Hence required equations are,
7x-y+8=0
(or)
x+7y-6=0

Answer 2

Step-by-step explanation:

I hope you understand because questions said clock wise