Dear Student, A3,1,B1,5,C4,2Using mid point formula, P = 1+42,5+22= 52,72Also using section formula, m:n = 3:1, C4,2 and A 3,1Q = 9+43+1,3+21+3= 134,54Also, m:n= 1:3 and A3,1,B1,5R = 9+13+1,3+51+3= 52,2Now line AP = line containing point 3,1, 52,72y - 1 =72-152-3 x-3y-1 = -x+3 ⇒ x+y-4=0 1Now line BQ = line containing point 1,5, 134,54y - 5 =54-5134-1 x-19y-5 = -15x+15 ⇒ 15x+9y-60=0 2Now line CR = line containing point 4,2, 52,2y - 2=2-252-4 x-4y-2=0 3Now if they are concurrent then there will be a common solution From equation 1 and 3 y = 2 and x = 2 but it is not satisfying equation 2 so they are not concurrentRegards

Pls solve 9... thanx

Dear Student, A (3, 1), B (1, 5), C (4, 2) Using mid point formula, P = (\frac{1 + 4}{2}, \frac{5 + 2}{2}) = (\frac{5}{2}, \frac{7}{2}) Also using section formula, m : n = 3 : 1, C (4, 2) and A (3, 1) Q = (\frac{9 + 4}{3 + 1}, \frac{3 + 2}{1 + 3}) = (\frac{13}{4}, \frac{5}{4}) Also, m : n = 1 : 3 and A (3, 1), B (1, 5) R = (\frac{9 + 1}{3 + 1}, \frac{3 + 5}{1 + 3}) = (\frac{5}{2}, 2) Now line AP = line containing point (3, 1), (\frac{5}{2}, \frac{7}{2}) y - 1 = \frac{\frac{7}{2} - 1}{\frac{5}{2} - 3} (x - 3) y - 1 = - x + 3 \Rightarrow x + y - 4 = 0 . . . . . (1) Now line BQ = line containing point (1, 5), (\frac{13}{4}, \frac{5}{4}) y - 5 = \frac{\frac{5}{4} - 5}{\frac{13}{4} - 1} (x - 1) 9 (y - 5) = - 15 x + 15 \Rightarrow 15 x + 9 y - 60 = 0 . . . . . (2) Now line CR = line containing point (4, 2), (\frac{5}{2}, 2) y - 2 = \frac{2 - 2}{\frac{5}{2} - 4} (x - 4) y - 2 = 0 . . . . . (3) Now if they are concurrent then there will be a common solution From equation 1 and 3 y = 2 and x = 2 but it is not satisfying equation 2 so they are not concurrent Regards

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