Question

Find the area of the triangle whose vertices are
(1) (2, 3) (-1,0), (2,-4)
(in) (0,0), (3,0) and (0,2)
(-5, -1), (3.-5. (32)​

Answer 1

ǫᴜᴇsᴛɪᴏɴ:

• Find the area of triangle whose vertices are
• (2,3)(-1,0)(2,-4)
• (-5,-1(3,-5)(3,2)

sᴏʟᴜᴛɪᴏɴ:

i) (2,3)(-1,0),(2,-4)

Let the two points be A(2,3) ,B(-1,0),C(2,-4)

Required formula for Area of triangle ABC = $\frac{1}{2} [ x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2}) ]$

$x_{1} = 2, y_{1} = 3$

$x _{2} = - 1,y_{2} = 0$

$x_{3} = 2,y_{3} = - 4$

Area of triangle ABC = $\frac{1}{2}$ [2(0-(-4))+(-1)(-4-3)+2(3-0)]

$= \frac{1}{2}$ [2(4) + (-1)(-7) + 2(3)]

$= \frac{1}{2}$ [8 + 7 + 6]

$= \frac{1}{2} \times 21$

= 10.5 square units

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ii) (-5,-1)(3,-5)(5,2)

Let the two points be A(-5,-1),B(3,-5),C(5,2)

$x_{1} = - 5,y_{1} = - 1$

$x _{2} = 3, y_{2} = - 5$

$x_{3} = 5,y_{3} = 2$

Area of triangle ABC = $\frac{1}{2}$ [(-5)(-5-2) + 3 (2-(-1)) + 5(-1-(-5))]

$= \frac{1}{2}$ [-5(-7) + 3(2+1) + 5(-1+5) ]

$= \frac{1}{2}$ [-5(-7) + 3(3) + 5(4) ]

$= \frac{1}{2}$ [35 + 9 + 20 ]

$= \frac{1}{2} \times 64$

= 32 square units

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