A is the point of intersection of the lines 3x+y−4=0 and x−y=0. If a line having negative slope makes an angle of 45∘ with the line x−3y+5=0 and passes through A then its equation is
To find point A, we need to solve the equations: ‌3x+y−4=0 ‌x−y=0 The second equation gives us y=x. Substitute y=x into the first equation: ‌3x+x−4=0 ‌4x−4=0 ‌4x=4 ‌x=1 Substituting x=1 into y=x, we get y=1. Thus, the coordinates of A are (1,1). Next, consider the line x−3y+5=0 which has a slope of ‌
1
3
. Let m be the slope of the required line. Since the line makes an angle of 45∘ with the given line and has a negative slope, we use the angle formula: tan‌45∘=|‌
‌
1
3
−m
1+‌
1
3
â‹…m
| Since tan‌45∘=1, we have: 1=|‌
‌
1−3m
3
3+m
3
| This simplifies to: |3+m|=|1−3m| Solving 3+m=±(1−3m) gives two possibilities: ‌3+m=1−3m ‌4m=−2‌‌⇒‌‌m=−‌
1
2
‌3+m=−1+3m ‌−2m=−4‌‌⇒‌‌m=2 Since the slope must be negative, we choose m=−‌
1
2
. The equation of the line using point-slope form is: y−1=−‌
1
2
(x−1) Simplifying: ‌2(y−1)=−(x−1) ‌2y−2=−x+1 ‌x+2y=3 Thus, the required equation of the line is: x+2y=3