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The area of triangle formed by the points $(0, 3)$ , (1,2) and $(\frac{15}{7}, \frac{26}{7})$ is:

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9

sq. units

\frac{10}{7}

sq. units

4

sq. units

\frac{16}{7}

sq. units

answer is B.

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Detailed Solution

Given points are (0,3), (1,2) and

(\frac{15}{7}, \frac{26}{7})

.
Let the points be A(0,3), B(1,2) and C

(\frac{15}{7}, \frac{26}{7})

.
If

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

are the coordinates of a triangle then,
Area of

ΔABC = \frac{1}{2} | (x_{1} y_{2} - x_{2} y_{1}) + (x_{2} y_{3} - x_{3} y_{2}) + (x_{3} y_{1} - x_{1} y_{3}) |

\Rightarrow Area of △ ABC = \frac{1}{2} | (0 - 3) + (\frac{26}{7} - \frac{30}{7}) + (\frac{45}{7} - 0) |

\Rightarrow Area of △ ABC = \frac{1}{2} | - 3 - \frac{4}{7} + \frac{45}{7} |

\Rightarrow Area of △ ABC = \frac{1}{2} \times \frac{20}{7}

\Rightarrow Area of △ ABC = \frac{10}{7} sq . units

Thus, area of triangle is

\frac{10}{7} sq . units

.
Therefore, option (2) is correct.

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