The co-ordinate of the points A,B,C …
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Sia ? 6 years, 6 months ago
Area of {tex}\triangle{/tex}ABC
{tex}= \frac { 1 } { 2 }{/tex} [6(5 + 2) - 3(-2 - 3) + 4(3 - 5)
{tex}= \frac { 1 } { 2 }{/tex} [42 + 15 - 8]
{tex}= \frac { 49 } { 2 }{/tex} sq. units
Area of {tex}\triangle{/tex}PBC
{tex}= \frac { 1 } { 2 }{/tex} [x(- 2 - 5) + 4(5 - y) - 3(y + 2)
{tex}= \frac { 1 } { 2 }{/tex} [-7x + 20 - 4y - 3y - 6]
{tex}= \frac { 1 } { 2 }{/tex} [-7x - 7y + 14]
{tex}= \frac { 1 } { 2 }{/tex} [-7(x + y - 2)]
{tex}= \frac { 7 } { 2 }{/tex} [x + y - 2] sq. units
Now, {tex}\frac { A r ( \triangle P B C ) } { A r ( \triangle A B C ) }{/tex}{tex}= \frac { \frac { 7 } { 2 } | x + y - 2 | } { \frac { 49 } { 2 } }{/tex}
{tex}= \frac { | x + y - 2 | } { 7 } = \frac { | x + y - 2 | } { | 7 | }{/tex}
{tex}= \left| \frac { x + y - 2 } { 7 } \right|{/tex}
Hence, proved
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