
The equation of the line bisecting the line segment joining the points \[\left( {a,{\rm{ }}b} \right)\]and \[\left( {a\prime ,\;b\prime } \right)\]at right angle, is
A)\[2(a - a\prime )x + 2(b - b\prime )y = a2 + b2 - a\prime 2 - b\prime 2\]
B) \[\;\;(a - a\prime )x + (b - b\prime )y = a2 + b2 - a\prime 2 - b\prime 2\]
C) \[2(a - a\prime )x + 2(b - b\prime )y = a\prime 2 + b\prime 2 - a2 - b2\]
D) None of these
Answer
162.3k+ views
Hint: First calculate the midpoints of line segment AB and CD through which required straight line passes. Midpoint is a point which divides the line in equal proportion. Bisection means the point divides the line in two equal parts. Use coordinates of the given points in order to find the slope of the required straight line.
Formula Used:Coordinates of midpoints is
Midpoint = \[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
Where
\[\left( {{x_1},{y_{1\;}}} \right),\;\left( {{x_2},{y_2}} \right)\]Are coordinates of two points on straight line
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Where
\[\left( {{x_1},\;{y_1}} \right)\]is coordinates through which line is passes
m is the slope of straight line
Complete step by step solution:Given: coordinates of two points
Midpoint=\[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
Midpoint=\(\left( {\dfrac{{\left( {a + a'} \right)}}{2},\;\dfrac{{\left( {b + b'} \right)}}{2}} \right)\)
According to question required line is passes through point having coordinate:
\(\left( {\dfrac{{\left( {a + a'} \right)}}{2},\;\dfrac{{\left( {b + b'} \right)}}{2}} \right)\)
According to question lines bisect at right angle:
Slope of required straight line
\[m = \dfrac{{ - 1}}{{\left( {\dfrac{{b' - b}}{{a' - a}}} \right)}} = \dfrac{{ - \left( {a' - a} \right)}}{{b' - b}}\]
Equation of required Straight line:
\[y - \] \[\dfrac{{\left( {b + b'} \right)}}{2}\]=\[\dfrac{{ - \left( {a' - a} \right)}}{{b' - b}}\]\[(x - \dfrac{{a + a'}}{2})\]
On simplification we get the required equation of straight line
\[2(b - b\prime )y + 2(a - a\prime )x - b2 + b\prime 2 - a2 + a\prime 2 = 0.\]
Option ‘A’ is correct
Note:Do not use the equation of line in any other form because it will become very difficult to find the equation of lines and sometimes one may not find the equation of line by using the other equation of lines. Slope of line is found using the coordinates of two points through which lines pass. Use the one point formula of a straight line because two points through which line passes are given indirectly in the question.
Formula Used:Coordinates of midpoints is
Midpoint = \[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
Where
\[\left( {{x_1},{y_{1\;}}} \right),\;\left( {{x_2},{y_2}} \right)\]Are coordinates of two points on straight line
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Where
\[\left( {{x_1},\;{y_1}} \right)\]is coordinates through which line is passes
m is the slope of straight line
Complete step by step solution:Given: coordinates of two points
Midpoint=\[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
Midpoint=\(\left( {\dfrac{{\left( {a + a'} \right)}}{2},\;\dfrac{{\left( {b + b'} \right)}}{2}} \right)\)
According to question required line is passes through point having coordinate:
\(\left( {\dfrac{{\left( {a + a'} \right)}}{2},\;\dfrac{{\left( {b + b'} \right)}}{2}} \right)\)
According to question lines bisect at right angle:
Slope of required straight line
\[m = \dfrac{{ - 1}}{{\left( {\dfrac{{b' - b}}{{a' - a}}} \right)}} = \dfrac{{ - \left( {a' - a} \right)}}{{b' - b}}\]
Equation of required Straight line:
\[y - \] \[\dfrac{{\left( {b + b'} \right)}}{2}\]=\[\dfrac{{ - \left( {a' - a} \right)}}{{b' - b}}\]\[(x - \dfrac{{a + a'}}{2})\]
On simplification we get the required equation of straight line
\[2(b - b\prime )y + 2(a - a\prime )x - b2 + b\prime 2 - a2 + a\prime 2 = 0.\]
Option ‘A’ is correct
Note:Do not use the equation of line in any other form because it will become very difficult to find the equation of lines and sometimes one may not find the equation of line by using the other equation of lines. Slope of line is found using the coordinates of two points through which lines pass. Use the one point formula of a straight line because two points through which line passes are given indirectly in the question.
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