Find the coordinates of the vertices of a triangle- the equations of whose sides are-i- x - y minus- 4 - 0- 2x minus- y - 3 - 0 and x minus- 3y - 2 - 0-ii- y -t1 - t2- - 2x - 2a t1t2- y -t2 - t3- - 2x - 2a t2t3 and- y -t3 - t1- - 2x - 2a t1t3-

(i) x + y minus; 4 = 0, 2x minus; y + 3 = 0 and x minus; 3y + 2 = 0x + y minus; 4 = 0 nbsp; nbsp; nbsp; nbsp; nbsp; ... (1)2x minus; y ...

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Byju's Answer

Standard XII

Mathematics

Quadratic Polynomial

Find the coor...

Question

Find the coordinates of the vertices of a triangle, the equations of whose sides are
(i) x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0
(ii) y (t₁ + t₂) = 2x + 2a t₁t₂, y (t₂ + t₃) = 2x + 2a t₂t₃ and, y (t₃ + t₁) = 2x + 2a t₁t₃.

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Solution

(i) x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0

x + y − 4 = 0 ... (1)

2x − y + 3 = 0 ... (2)

x − 3y + 2 = 0 ... (3)

Solving (1) and (2) using cross-multiplication method:

$\frac{x}{3 - 4} = \frac{y}{- 8 - 3} = \frac{1}{- 1 - 2} \Rightarrow x = \frac{1}{3}, y = \frac{11}{3}$

Solving (1) and (3) using cross-multiplication method:

$\frac{x}{2 - 12} = \frac{y}{- 4 - 2} = \frac{1}{- 3 - 1} \Rightarrow x = \frac{5}{2}, y = \frac{3}{2}$

Similarly, solving (2) and (3) using cross-multiplication method:

$\frac{x}{- 2 + 9} = \frac{y}{3 - 4} = \frac{1}{- 6 + 1} \Rightarrow x = - \frac{7}{5}, y = \frac{1}{5}$

Hence, the coordinates of the vertices of the triangle are $(\frac{1}{3}, \frac{11}{3})$ , $(\frac{5}{2}, \frac{3}{2})$ and $(- \frac{7}{5}, \frac{1}{5})$ .

(ii) y (t₁ + t₂) = 2x + 2a t₁t₂, y (t₂ + t₃) = 2x + 2a t₂t₃ and y (t₃ + t₁) = 2x + 2a t₁t₃

2x − y (t₁ + t₂) + 2a t₁t₂ = 0 ... (1)

2x − y (t₂ + t₃) + 2a t₂t₃ = 0 ... (2)

2x − y (t₃ + t₁) + 2a t₁t₃ = 0 ... (3)

Solving (1) and (2) using cross-multiplication method:

$\frac{x}{- 2 a t_{2} t_{3} (t_{1} + t_{2}) + 2 a t_{1} t_{2} (t_{2} + t_{3})} = \frac{y}{4 a t_{1} t_{2} - 4 a t_{2} t_{3}} = \frac{1}{- 2 (t_{2} + t_{3}) + 2 (t_{1} + t_{2})} \Rightarrow \frac{x}{2 a {t_{2}}^{2} (t_{1} - t_{3})} = \frac{y}{4 a t_{2} (t_{1} - t_{3})} = \frac{1}{2 (t_{1} - t_{3})} \Rightarrow x = a {t_{2}}^{2}, y = 2 a t_{2}$

Solving (1) and (3) using cross-multiplication method:

$\frac{x}{- 2 a t_{1} t_{3} (t_{1} + t_{2}) + 2 a t_{1} t_{2} (t_{3} + t_{1})} = \frac{y}{4 a t_{1} t_{2} - 4 a t_{1} t_{3}} = \frac{1}{- 2 (t_{3} + t_{1}) + 2 (t_{1} + t_{2})} \Rightarrow \frac{x}{2 a {t_{1}}^{2} (t_{2} - t_{3})} = \frac{y}{4 a t_{1} (t_{2} - t_{3})} = \frac{1}{2 (t_{2} - t_{3})} \Rightarrow x = a {t_{1}}^{2}, y = 2 a t_{1}$

Similarly, solving (2) and (3) using cross-multiplication method:

$\frac{x}{- 2 a t_{1} t_{3} (t_{2} + t_{3}) + 2 a t_{2} t_{3} (t_{3} + t_{1})} = \frac{y}{4 a t_{2} t_{3} - 4 a t_{1} t_{3}} = \frac{1}{- 2 (t_{3} + t_{1}) + 2 (t_{2} + t_{3})} \Rightarrow \frac{x}{2 a {t_{3}}^{2} (t_{2} - t_{1})} = \frac{y}{4 a t_{3} (t_{2} - t_{1})} = \frac{1}{2 (t_{2} - t_{1})} \Rightarrow x = a {t_{3}}^{2}, y = 2 a t_{3}$

Hence, the coordinates of the vertices of the triangle are $(a {t_{1}}^{2}, 2 a t_{1})$ , $(a {t_{2}}^{2}, 2 a t_{2})$ and $(a {t_{3}}^{2}, 2 a t_{3})$ .

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Find the coordinates of the vertices of a triangle, the equations of whose sides are (i) x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0 (ii) y (t1 + t2) = 2x + 2a t1t2, y (t2 + t3) = 2x + 2a t2t3 and, y (t3 + t1) = 2x + 2a t1t3.

Find the coordinates of the vertices of a triangle, the equations of whose sides are
(i) x + y − 4 = 0, 2x − y + 3 = 0 and x − 3y + 2 = 0
(ii) y (t₁ + t₂) = 2x + 2a t₁t₂, y (t₂ + t₃) = 2x + 2a t₂t₃ and, y (t₃ + t₁) = 2x + 2a t₁t₃.