A triangle has vertices A - x 1- y 1 for i -1-2-3- Then the determinant Delta - Vegin vmatrix x -2- x -3- y -2- y -3- y -1 y -2- y -3- x -1 x -2- x -3 1x -3- x -1- y -3- y -1- y -2 y -3- y -1- x -2 x -3- x -1 1x-1-x-2- y-1-y-2- y-3y-1- y -2- x -3x-1-x-2 end vmatrix -0 - means A- medians of the triangle A 1- A 2- A 3 are concurrent-B- the triangle A 1- A 2- A 3 is right angled at A 3-C- the triangle A 1- A 2- A 3 is an equilateral triangle-D- altitudes of the triangle A 1- A 2- A 3 are concurrent-

The correct option is D altitudes of the triangle A1,A2,A3 are concurrent. If altitudes of a triangle are concurrent at H (0, 0), then AH ⊥ A2, A3 y_1 0/x_1 ...

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Byju's Answer

Standard XII

Mathematics

Area of Triangle

A triangle ha...

Question

A triangle has vertices A, $(x_{1}, y_{1})$ for i=1,2,3. Then the determinant

$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{3} & y_{2} - y_{3} & y_{1} (y_{2} - y_{3}) + x_{1} (x_{2} - x_{3}) x_{3} - x_{1} & y_{3} - y_{1} & y_{2} (y_{3} - y_{1}) + x_{2} (x_{3} - x_{1}) x_{1} - x_{2} & y_{1} - y_{2} & y_{3} (y_{1} - y_{2}) + x_{3} (x_{1} - x_{2}) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0 m e a n s$

medians of the triangle A₁,A₂,A₃ are concurrent;

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the triangle A₁,A₂,A₃ is right angled atA₃;

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the triangle A₁,A₂,A₃ is an equilateral triangle;

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altitudes of the triangle A₁,A₂,A₃are concurrent.

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Solution

The correct option is D
altitudes of the triangle A₁,A₂,A₃are concurrent.

If altitudes of a triangle are concurrent at H (0, 0), then AH ⊥ A₂, A₃

$\frac{y_{1} - 0}{x_{1} - 0} = - 1$ ⇒y₁(y₂-y₃)+x₁(x₂-x₃)=0 ⇒∆=0

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Similar questions

Q. Let

A_{1}, A_{2}, A_{3}

be regions in the xy-plane defined by

A_{1} = {(x, y) : x^{2} + 2 y^{2} \leq 1}

A_{2} = {(x, y) : | x |^{3} + 2 \sqrt{2} | y |^{3} \leq 1}

A_{3} = {(x, y) : m a x (| x |, \sqrt{2} | y |) \leq 1}

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If the lines a_{1} x + b_{1} y + 1 = 0, a_{2} x + b_{2} y + 1 = 0 and a_{3} x + b_{3} y + 1 = 0 are concurrent, then the points (a_{1}, b_{1}), (a_{2}, b_{2}) and (a_{3}, b_{3}) will be collinear .

Q. Let

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A_{1} = {(x, y) : x \geq 0, y \geq 0, 2 x + 2 y - x^{2} - y^{2} > 1 > x + y}, A_{2} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{2} + y^{2}}, A_{3} = {(x, y) : x \geq 0, y \geq 0, x + y > 1 > x^{3} + y^{3}} .

Denote by

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Q. An equilateral triangle has each of its sides of length

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{∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣}^{2}

Q. If the coordinates of the vertices of an equilateral triangle with sides of length

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and

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{∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣}^{2} =

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A triangle has vertices A, (x1,y1) for i=1,2,3. Then the determinant Δ=∣∣ ∣ ∣∣x2−x3y2−y3y1(y2−y3)+x1(x2−x3)x3−x1y3−y1y2(y3−y1)+x2(x3−x1)x1−x2y1−y2y3(y1−y2)+x3(x1−x2)∣∣ ∣ ∣∣=0 means

The correct option is D altitudes of the triangle A1,A2,A3 are concurrent. If altitudes of a triangle are concurrent at H (0, 0), then AH ⊥ A2, A3 y1−0x1−0=−1⇒y1(y2-y3)+x1(x2-x3)=0 ⇒∆=0

The correct option is D
altitudes of the triangle A₁,A₂,A₃are concurrent.

If altitudes of a triangle are concurrent at H (0, 0), then AH ⊥ A₂, A₃

$\frac{y_{1} - 0}{x_{1} - 0} = - 1$ ⇒y₁(y₂-y₃)+x₁(x₂-x₃)=0 ⇒∆=0