Straight Lines and Pair of Straight Lines

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 ⋅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