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If the co-ordinates of points P,Q & R satisfy the relation xy=c 2 , show that the orthocentre of the triangle PQR also satisfies this relation If the co-ordinates of points P,Q & R satisfy the relation xy=c2, show that the orthocentre of the triangle PQR also satisfies this relation
If the co-ordinates of points P,Q & R satisfy the relation xy=c2, show that the orthocentre of the triangle PQR also satisfies this relation
THERE IS ONE LONG METHOD AND SHORT METHOD ... LONG METHOD: CONSIDER THREE POINTS OF P,Q,R AS (X1,Y1),(X2,Y2),(X3,Y3) NOW SINCE ORTHOCENTER IS THE INTERSECTION OF ALTITUDES SO FIND THE EQUATION OF ALTITUDES . AND FIND THE INTERSECTING POINTS AND PUT THE POINTS IN THE EQUATION WHICH IS GIVEN AND SEE IF IT SATISFIES .... EQUATION OF THE ALTITUDE WILL BE FOUND OUT BY YOU WILL KNOW THE SLOPE OS THE SIDES SINCE YOU HAVE THE POINTS AND ANY LINE PERPENDICULAR TO IT WILL HAVE SLOPE EQUAL TO -1/THE SLOPE OF THE LINE THEN YOU CAN FIND EQUATION BY POINT FORM AND PROCEED....... .......................................................................................................................................................................... SHORT METHOD..... IF YOU LOOK AT THE EQUATION CAREFULLY YOU WILL (XY=C^2) NOTICE THAT THIS EQUATION IS THE LOCUS OF POINTS ON A RECTANGULAR HYPERBOLA ......... SO THERE IS ONE PROPERTY THAT IF THREE SIDES OF THE TRIANGLE LIES ON A RECTANGULAR HYPERBOLA THEN THE ORTHOCENTER OF THE TRIANGLE WILL ALSO LIE ON THE RECTANGULAR HYPERBOLA
THERE IS ONE LONG METHOD AND SHORT METHOD ...
LONG METHOD: CONSIDER THREE POINTS OF P,Q,R AS (X1,Y1),(X2,Y2),(X3,Y3) NOW SINCE ORTHOCENTER IS THE INTERSECTION OF ALTITUDES SO FIND THE EQUATION OF ALTITUDES . AND FIND THE INTERSECTING POINTS AND PUT THE POINTS IN THE EQUATION WHICH IS GIVEN AND SEE IF IT SATISFIES .... EQUATION OF THE ALTITUDE WILL BE FOUND OUT BY
YOU WILL KNOW THE SLOPE OS THE SIDES SINCE YOU HAVE THE POINTS AND ANY LINE PERPENDICULAR TO IT WILL HAVE SLOPE EQUAL TO -1/THE SLOPE OF THE LINE THEN YOU CAN FIND EQUATION BY POINT FORM AND PROCEED.......
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SHORT METHOD..... IF YOU LOOK AT THE EQUATION CAREFULLY YOU WILL (XY=C^2) NOTICE THAT THIS EQUATION IS THE LOCUS OF POINTS ON A RECTANGULAR HYPERBOLA ......... SO THERE IS ONE PROPERTY THAT IF THREE SIDES OF THE TRIANGLE LIES ON A RECTANGULAR HYPERBOLA THEN THE ORTHOCENTER OF THE TRIANGLE WILL ALSO LIE ON THE RECTANGULAR HYPERBOLA
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