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1) 95/6
2) 285/2
3) 190/3
4) 285
Solution:
We know the centroid of a triangle is {(x1+x2+x3)/3, (y1+y2+y3)/3} Here centroid is (0, 0) and two vertices are (-8, 7) and (9, 4). So (x1 – 8 + 9)/3 = 0 and (y1 + 7 + 4) = 0 => x1 + 1 = 0 and y1 + 11 = 0 => x1 = -1, y1 = -11 So the third vertex is (-1, -11) Area of triangle ∆ = \(\begin{array}{l}\frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1\\ x_{2}&y_{2} & 1\\ x_{3}& y_{3} & 1 \end{vmatrix}\end{array} \) = \(\begin{array}{l}\frac{1}{2}\begin{vmatrix} -1 & -11 & 1\\ -8&7 & 1\\ 9& 4 & 1 \end{vmatrix}\end{array} \)
= |(½)[ -1(7 – 4) + 11(-8 -9) + 1(-32 – 63)]|
= |(½)[ -3 – 187 – 95]|
= |-285/2|
= 285/2
Hence option (2) is the answer.
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5 (2)
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We have to find the co-ordinates of the third vertex of the given triangle. Let the co-ordinates of the third vertex be ( x , y) .
The co-ordinates of other two vertices are (4,−3) and (−9, 7)
The co-ordinate of the centroid is (1, 4)
We know that the co-ordinates of the centroid of a triangle whose vertices are `(x_1 ,y_1 ) , (x_2,y_2),(x_3,y_3)` is
`((x_1+x_2 +x_3)/3 , (y_1 + y_2+y_3)/3)`
So,
`(1 , 4) = ((x+4-9)/3 , (y-3+7)/3)`
Compare individual terms on both the sides- `(x - 5)/3 = 1`
So,
x = 8
Similarly,
`(y+ 4 )/3 = 4`
So,
y = 8
So the co-ordinate of third vertex is (8, 8)
In general if `A (x_1 , y_1) ;B(x_2 , y_2 ) ;C(x_3 , y_3)` are non-collinear points then are of the triangle formed is given by-,
`ar (Δ ABC ) = 1/2 |x_1(y_2 - y_3 ) +x_2 (y_3 - y_1) + x_3 (y_1 - y_2)|`
So,
`ar (ΔABC ) = 1/2 |4(7-8)-9(8+3)+8(-3-7)|`
`= 1/2 | -4-99-80|`
`= 183/2`