WebIf the lines x =ay +b, z= cy +d and x = a'y +b', z=c' y +d' are ⊥, then (A) ab' +bc' +1 =0 (B) bb' +cc' +1=0 (C) aa' +cc' +1=0 (D) aa' +bb' +1=0. Ch. Tardigrade ... WebAnswer in Brief If the lines x + ay + a = 0, bx + y + b = 0 and cx + cy + 1 = 0 are concurrent, then write the value of 2abc − ab − bc − ca. Advertisement Remove all ads Solution The given lines are x + ay + a = 0 ... (1) bx + y + b = 0 ... (2) cx + cy + 1 = 0 ... (3) It is given that the lines (1), (2) and (3) are concurrent.

If the lines x+ay=a=0, bx+y+b=0, cx+cy+1=0 (a,b,c being distinct a

WebIf the lines x + ay + a = 0, bx + y + b = 0 and cx + cy + 1 = 0 are concurrent, then the value of (a/a -1) + (b/b - 1) + (c/c -1) is Tardigrade Mathematics Q. Let a,b and c be distinct and none of them is equal to 1. If the lines x + ay + a = 0,bx + y + b = 0 and cx + cy + 1 = 0 are concurrent, then the value of a−1a + b−1b + c−1c is WebSince the lines x +2ay+ a = 0 x+ 3by+b = 0 and x+4cy +c = 0 are concurrent ∴ ∣∣1 1 1 2a 3b 4c a b c∣∣ = 0 Applying R2 → R2 − R1 and R3 → R3 −R1, we get ∣∣1 0 0 2a 3b −2a 4c −2a a b −a c −a∣∣ = 0 (3b− 2a) (c− a)−(4c −2a) (b −a) = 0 ⇒ −3bc− 3ab−2ac +2a2 −4bc − 4ac +2ab−2a2 = 0 ⇒ −bc −ab+ 2ac = 0 ⇒ 2ac = ab+bc ⇒ b1 = a1 + c1 Hence, a, b, c are in HP. frederick\\u0027s san antonio

ordinary differential equations - solve $y

WebThe Straight Lines VSAQ Solutions WebLeague of Legends, Twitch, poodle 26 views, 3 likes, 2 loves, 3 comments, 7 shares, Facebook Watch Videos from Syrèn: Let's Play - League of Legends... WebIf you have y ′ ( x) = a y ( x) + b ( x), for constant a, then the simpliest way is to multiply the equation by e − a x (nonzero everywhere) and deduce that ( y ( x) e − a x) ′ = e − a x b ( x), which implies that y ( x) e − a x − y ( 0) = ∫ 0 x e − a s b ( s) d s. Finally, y ( x) = y ( 0) e a x + ∫ 0 x e − a ( x − s) b ( s) d s. Share blind melon chitlin going downtown