The vertices of a triangle are - - and - Find the area of the triangle-

Given, vertices of the triangle are (2,3), ( 1,0) nbsp;and ( 2,4). ∴ Area of the triangle =12[x_1(y_2 y_3)+x_2(y_3 y_1)+x_3(y_1 y_2)] =12[2{ 0 ( 4)}+( 1){ 4 ...

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Byju's Answer

Standard X

Mathematics

Area of a Triangle Using Coordinate of Vertices

The vertices ...

Question

The vertices of a triangle are (𝟐,𝟑), (−𝟏,𝟎) and (𝟐,−𝟒). Find the area of the triangle.

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Solution

Given, vertices of the triangle are (2,3), (-1,0) and (-2,4).
$∴$ Area of the triangle $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$= \frac{1}{2} [2 {0 - (- 4)} + (- 1) {- 4 - 3)} + 2 {3 - 0}]$
$= \frac{1}{2} [8 + 7 + 6]$
$= \frac{21}{2}$ sq. units

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The vertices of a triangle are (𝟐,𝟑), (−𝟏,𝟎) and (𝟐,−𝟒). Find the area of the triangle.

Given, vertices of the triangle are (2,3), (-1,0) and (-2,4). ∴ Area of the triangle =12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)] =12[2{0−(−4)}+(−1){−4−3)}+2{3−0}] =12[8+7+6] =212 sq. units

Given, vertices of the triangle are (2,3), (-1,0) and (-2,4).
$∴$ Area of the triangle $= \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$= \frac{1}{2} [2 {0 - (- 4)} + (- 1) {- 4 - 3)} + 2 {3 - 0}]$
$= \frac{1}{2} [8 + 7 + 6]$
$= \frac{21}{2}$ sq. units