Find the area of -ABC with vertices A-1- minus-3- B-4- minus-3- and C- minus-9- 7-

The given points are nbsp;A1, #160; 3, #160;B4, #160; 3 #160;and #160;C 9, #160;7.Here, nbsp;x1=1, #160;y1= 3, #160;x2=4, #160;y2= 3 #16 ...

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Standard XII

Mathematics

Area of Triangle with Coordinates of Vertices Given

Find the area...

Question

Find the area of ∆ABC with vertices A(1, −3), B(4, −3) and C(−9, 7).

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Solution

The given points are $A (1, - 3), B (4, - 3) and C (- 9, 7)$ .
Here, $(x_{1} = 1, y_{1} = - 3), (x_{2} = 4, y_{2} = - 3) and (x_{3} = - 9, y_{3} = 7)$
Therefore,
$Area of ∆ A B C = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$

$= \frac{1}{2} |1 (- 3 - 7) + 4 (7 + 3) - 9 (- 3 + 3)| = \frac{1}{2} |- 10 + 40 + 0| = (\frac{1}{2} \times 30) = 15 sq units$

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Find the area of ∆ABC with vertices A(1, −3), B(4, −3) and C(−9, 7).

The given points are A1, -3, B4, -3 and C-9, 7.Here, x1=1, y1=-3, x2=4, y2=-3 and x3=-9, y3=7Therefore,Area of ∆ABC=12x1y2-y3+x2y3-y1+x3y1-y2 =121-3-7+47+3-9-3+3=12-10+40+0=12×30=15 sq units