A line passing through the point P(1, 2) meets the line x+y=7 at the distance of 3 units from P. Then the slope of this line satisfies the equation

An equation of line through (1, 2) is y – 2 = m(x – 1) A point on this line is of the form Q(x, mx – m + 2). This point will lie on the line x + y = 7 if x+mx−m+2=7⇒x=m+5m+1 As PQ=3,PQ2=9⇒(x−1)2+(mx−m)2=9⇒(x−1)21+m2=9⇒16m2+16=9m2+18m+9⇒7m2−18m+7=0⇒m satisfies the equation 7x2−18x+7=0