AP EAPCET 20-Aug-2021 Shift 2 Paper

Let point (1,2) be denoted as P(1,2)
Given line, 3x−y=7 have slope 3.
Given, PQ||3x−y=7
⇒ Slope of PQ=3
Equation of line PQ which passes through (1,2) and have slope 3 is
y−2=3(x−1)
⇒y−3x+1=0
Then point of intersection of PQ and x+y+5=0 is obtained by solving
x+y=−5
3x−y=1
Solve, we get x=−1,y=−4
∴ Point Q(−1,−4)
Distance‌PQ=√(−1−1)2+(−4−2)2 =√4+36=√40