If x 1- y 1- x 2- y 2 and x 3- y 3 are the vertices of a triangle whose area is - k - square units- then x - 1 - y - 1 - 4 llx2 - y2 - 411 x- 3 - y- 3 - 4lend vmatrix -2 isA- 32 k 2B- 16 k 2- 64 k 2D- 48 k 2

The correct option is C 64k^2 nbsp; nbsp; nbsp;Area of a triangle with vertices nbsp;(x_1, y_1), nbsp;(x_2, y_2) and nbsp; nbsp;(x_3, y_3) n ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

If x 1, y 1, ...

Question

If $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ are the vertices of a triangle whose area is ' $k$ ' square units, then ${∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 4 x_{2} & y_{2} & 4 x_{3} & y_{3} & 4 \end{matrix} ∣ ∣ ∣ ∣}^{2}$ is

32 k^{2}

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16 k^{2}

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64 k^{2}

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48 k^{2}

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Solution

The correct option is C $64 k^{2}$
Area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ is given by
$\frac{1}{2} ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣$

$\Rightarrow ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 2 k$

$\Rightarrow ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 4 x_{2} & y_{2} & 4 x_{3} & y_{3} & 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ = 4 \times 2 k = 8 k$

$∴ {∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 4 x_{2} & y_{2} & 4 x_{3} & y_{3} & 4 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣}^{2} = (8 k)^{2} = 64 k^{2}$

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Similar questions

Find the centroid of the triangle with vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ .

If the coordinates of the vertices of an equilateral triangle with sides of length 'a' are $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ , then
${∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 1 x_{2} & y_{2} & 1 x_{3} & y_{3} & 1 \end{matrix} ∣ ∣ ∣ ∣}^{2} = \frac{3 a^{4}}{4}$

Prove that the coordinates of the centroid of the triangle whose vertices are $(x_{1}, y_{1}), (x_{2}, y_{2}) a n d (x_{3}, y_{3})$ are $(\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$ and also, deduce that the medians of a triangles are concurrent.

Q. If

A (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3})

are the vertices of an equilateral triangle and

(x_{1} - 4)^{2} + (y_{1} - 5)^{2} = (x_{2} - 4)^{2} + (y_{2} - 5)^{2} = (x_{3} - 4)^{2} + (y_{3} - 5)^{2},

then the value of

(y_{1} + y_{2} + y_{3}) - (x_{1} + x_{2} + x_{3})

Q. If

A (x_{1}, y_{1}), B x_{2}, y_{2}), C (x_{3}, y_{3})

are the vertices of an equilateral triangle and

(x_{1} - 2)^{2} + (y_{1} - 3)^{2} = (x_{2} - 2)^{2} + (y_{2} - 3)^{2} = (x_{3} - 2)^{2} + (y_{3} - 3)^{2}

, then

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If (x1,y1),(x2,y2) and (x3,y3) are the vertices of a triangle whose area is 'k' square units, then ∣∣ ∣∣x1y14x2y24x3y34∣∣ ∣∣2 is

If $(x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3})$ are the vertices of a triangle whose area is ' $k$ ' square units, then ${∣ ∣ ∣ ∣ \begin{matrix} x_{1} & y_{1} & 4 x_{2} & y_{2} & 4 x_{3} & y_{3} & 4 \end{matrix} ∣ ∣ ∣ ∣}^{2}$ is