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