Q. 165.0( 1 Vote )

# The area of the t A. 1/2 sq. unit

B. 1 sq. unit

C. 2 sq. unit

D. None of these

Answer :

Given :

Equation 1: x = 3

Equation 2: y = 4

Equation 3: x = y

Equation 1 is a line parallel to y axis

Equation 2 is a line parallel to x axis

So Equation 1 & 2 are mutually perpendicular to each other.

Hence the triangle formed is a right angled triangle.

First we solve the three lines simultaneously by method of substitution and get the three points of intersection or three coordinates of the triangle.

Solving Equation 1 & 2 we get the coordinate ( 3, 4 ). Let this Coordinate name be P_{1}

Solving Equation 2 & 3 we get the coordinate ( 4 ,4 ). Let this Coordinate name be P_{2}

Solving Equation 3 & 1 we get the coordinate ( 3 ,3 ). Let this Coordinate name be P_{3}

We now use the formula for Area of a triangle through 3 given points

Area = *| x_{1} * (y_{2} – y_{3}) + x_{2} * (y_{3} – y_{1}) + x_{3} * (y_{1} – y_{2}) |

Where x_{1 ,}y_{1} are the coordinates of P_{1}

x_{2,} y_{2} are the coordinates of P_{2}

x_{3 ,}y_{3} are the coordinates of P_{3}

Area of the Given Triangle = *| 3* (4– 3) + 4 * (3 – 4) + 3* (4– 4) |

Area = *| 3* (1) + 4 * ( – 1) + 3* (0) |

Area = *| 3– 4 |⇒ Area = sq. units

__The Area of the triangle is__ sq. units

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