Find the area of triangle with …
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Sia ? 6 years, 6 months ago
Let E be the midpoint of AB.
{tex}\therefore \quad \frac { x + 1 } { 2 } = 2{/tex} or x = 3
and {tex}\frac { y + ( - 4 ) } { 2 } = - 1{/tex} or, y = 2
or, B(3, 2)
Let F be the mid-point of AC.Then,
{tex}0=\frac{x_1+1}{2}{/tex} or {tex}x_1=-1{/tex}
and {tex}\frac { y _ { 1 } + ( - 4 ) } { 2 }{/tex} = -1 or, y1 = 2
or, C= (-1, 2)
Now the co-ordinates are A(1, - 4), B(3,2), C (-1,2)
Area of triangle
{tex}= \frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right]{/tex}
{tex}= \frac { 1 } { 2 } [ 1 ( 2 - 2 ) + 3 ( 2 + 4 ) - 1 ( - 4 - 2 ) ]{/tex}
{tex}= \frac { 1 } { 2 } [ 0 + 18 + 6 ]{/tex}
= 12 sq units.
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