Let At1, 2at1+t13; Bt2, 2at2+t23; Ct3, 2at3+t33 be the given points Since, A, B and C are collinear, thenar∆ABC = 0⇒12t12at1+t131t22at2+t231t3 2at3+t331 = 0⇒t12at11t22at21t3 2at31 + t1t131t2t231t3 t331 = 0⇒2at1t11t2t21t3 t31 + t1t131t2t231t3 t331 = 0⇒2a × 0 + t1t131t2t231t3 t331 = 0 ∆1 = 0 as C1 and C2 are identical⇒ t1t131t2t231t3 t331 = 0 Apply R1 = R1 - R2 and R2 = R2 - R3, we gett1-t2t13-t230t2-t3t23-t330t3 t331 = 0 ⇒t1-t2t2-t31t12+t22+t1t201t22+t32+t3t20t3 t331 = 0⇒t1-t2t2-t3 t22+t32+t3t2 - t12-t22-t1t2 = 0⇒t1-t2t2-t3t3-t1t3+t1+t2t3-t1 = 0⇒t1-t2t2-t3t3-t1t1+t2+t3 = 0Since, t1, t2 and t3 are distinct, So, t1 ≠ t2 ; t2 ≠ t3 and t3 ≠ t1So, t1 + t2 + t3 = 0

Pls solve Q13

Pls solve Q13 r ombus The area of the triangle With vertices at (-4, -1), (1, 2), (4, -3) is (D) none of these (D) 30 The area ofthe triangle With vertices (A) Jab(r- -1) (B) lab(r - I-lab(r - 11, 12 13 Ifthe area of the triangle formed by the points (1, 2). (2, 3), (x , 4) is 40 sq.unitsthen xis (A) 1/2,2 (B) 2, 213 (D) 1/2, -1 The area of the quadrilateral formed by the points (1, 2), (2, —3), (—2, 4), (0, 5) is (A) 10 sq.unit (B) 15 sq.unit (C) 18 sq.unit (D) 20 sq.unit If tl,t2, and % are distinct, the points (tl,2at1 +ag)and (t3,2at3 +at3) are collinear if (A) tlt2t3 = 1 (B) tl+t2 +t3 = tit2t3 14. The area of the triangle formed by the line x sina+ ycosa sin2a and the coordinates axes is (D)2cos2u (C)2sin2a 4(Å) sin 2a (B) cos2a 15 The area of the triangle bounded by the straight line ax +bY+c=0, and the coordinate

Let A (t_{1}, 2 {at}_{1} + {t_{1}}^{3}); B (t_{2}, 2 {at}_{2} + {t_{2}}^{3}); C (t_{3}, 2 {at}_{3} + {t_{3}}^{3}) be the given points . Since, A, B and C are collinear, then ar (∆ ABC) = 0 \Rightarrow \frac{1}{2} |\begin{matrix} t_{1} & 2 {at}_{1} + {t_{1}}^{3} & 1 \\ t_{2} & 2 {at}_{2} + {t_{2}}^{3} & 1 \\ t_{3} & 2 {at}_{3} + {t_{3}}^{3} & 1 \end{matrix}| = 0 \Rightarrow |\begin{matrix} t_{1} & 2 {at}_{1} & 1 \\ t_{2} & 2 {at}_{2} & 1 \\ t_{3} & 2 {at}_{3} & 1 \end{matrix}| + |\begin{matrix} t_{1} & {t_{1}}^{3} & 1 \\ t_{2} & {t_{2}}^{3} & 1 \\ t_{3} & {t_{3}}^{3} & 1 \end{matrix}| = 0 \Rightarrow 2 a |\begin{matrix} t_{1} & t_{1} & 1 \\ t_{2} & t_{2} & 1 \\ t_{3} & t_{3} & 1 \end{matrix}| + |\begin{matrix} t_{1} & {t_{1}}^{3} & 1 \\ t_{2} & {t_{2}}^{3} & 1 \\ t_{3} & {t_{3}}^{3} & 1 \end{matrix}| = 0 \Rightarrow 2 a \times 0 + |\begin{matrix} t_{1} & {t_{1}}^{3} & 1 \\ t_{2} & {t_{2}}^{3} & 1 \\ t_{3} & {t_{3}}^{3} & 1 \end{matrix}| = 0 [∆_{1} = 0 as C_{1} and C_{2} are identical] \Rightarrow |\begin{matrix} t_{1} & {t_{1}}^{3} & 1 \\ t_{2} & {t_{2}}^{3} & 1 \\ t_{3} & {t_{3}}^{3} & 1 \end{matrix}| = 0 Apply R_{1} = R_{1} - R_{2} and R_{2} = R_{2} - R_{3}, we get |\begin{matrix} t_{1} - t_{2} & {t_{1}}^{3} - {t_{2}}^{3} & 0 \\ t_{2} - t_{3} & {t_{2}}^{3} - {t_{3}}^{3} & 0 \\ t_{3} & {t_{3}}^{3} & 1 \end{matrix}| = 0 \Rightarrow (t_{1} - t_{2}) (t_{2} - t_{3}) |\begin{matrix} 1 & {t_{1}}^{2} + {t_{2}}^{2} + t_{1} t_{2} & 0 \\ 1 & {t_{2}}^{2} + {t_{3}}^{2} + t_{3} t_{2} & 0 \\ t_{3} & {t_{3}}^{3} & 1 \end{matrix}| = 0 \Rightarrow (t_{1} - t_{2}) (t_{2} - t_{3}) [{t_{2}}^{2} + {t_{3}}^{2} + t_{3} t_{2} - {t_{1}}^{2} - {t_{2}}^{2} - t_{1} t_{2}] = 0 \Rightarrow (t_{1} - t_{2}) (t_{2} - t_{3}) [(t_{3} - t_{1}) (t_{3} + t_{1}) + t_{2} (t_{3} - t_{1})] = 0 \Rightarrow (t_{1} - t_{2}) (t_{2} - t_{3}) (t_{3} - t_{1}) (t_{1} + t_{2} + t_{3}) = 0 Since, t_{1}, t_{2} and t_{3} are distinct, So, t_{1} \neq t_{2}; t_{2} \neq t_{3} and t_{3} \neq t_{1} So, t_{1} + t_{2} + t_{3} = 0

View Full Answer