Write the formula for the area of the triangle having its vertices at (x₁, y₁), (x2, y₂) and (x₃, y₃).

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Solution

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

The area ‘A’ encompassed by three points, and is also given by the formula,

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

ABCD is a parallelogram with vertices A (X₁, Y₁) , B(X₂, Y₂) and C (X₃, Y₃). Then the coordinates of the fourth vertex D in terms of the coordinates of A, B and C are

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})

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} - 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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Write the formula for the area of the triangle having its vertices at (x1, y1), (x2, y2) and (x3, y3).

The formula for the area ‘A’ encompassed by three points, and is given by the formula, The area ‘A’ encompassed by three points, and is also given by the formula,

Write the formula for the area of the triangle having its vertices at (x₁, y₁), (x2, y₂) and (x₃, y₃).

The formula for the area ‘A’ encompassed by three points, and is given by the formula,

The area ‘A’ encompassed by three points, and is also given by the formula,