
A line meets x-axis and y-axis at the points A and B respectively. If the middle point of AB be \(\left( {{x_1},{y_1}} \right)\), then the equation of the line is
A. \({y_1}x + {x_1}y = 2{x_1}{y_1}\)
B. \[{x_1}x\; + {y_1}y = 2{x_1}{y_1}\]
C. \[{y_1}\;x + {x_1}y\; = {x_1}{y_1}\]
D. \[{x_1}x + {y_1}y = {x_1}{y_1}\]
Answer
232.8k+ views
Hint: First calculate the midpoints of line segment AB .Midpoint is a point which divides the line in equal proportion. Midpoint of line AB is given in question. Equate the midpoint calculated by midpoint formula to given midpoint. x and y intercept are indirectly given in question so in this question intercept form of straight line’s equation used.
Formula Used:Coordinates of midpoints of line 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 lines.
Equation of line is
\[\dfrac{x}{a} + \dfrac{y}{b} = 1\]
Where
a is x intercept of line
b is y intercept of line
Complete step by step solution:Given: Middle point of line AB is given by
Midpoint=\[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
According to question required line meets meets x-axis and y-axis at the points A and B respectively
Let coordinate of A= \[(a,0)\] and coordinate of B = \[(0,b)\]
Now apply midpoint formula
Midpoint =\[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
Midpoint=\[\left( {\dfrac{{\left( {a + 0} \right)}}{2},\;\dfrac{{\left( {0 + b} \right)}}{2}} \right)\]
Midpoint of line AB= \[\left( {\dfrac{a}{2},\;\dfrac{b}{2}} \right)\]
middle point of AB be \(\left( {{x_1},{y_1}} \right)\)
\({x_1} = \dfrac{a}{2}\;,\;{y_1} = \dfrac{b}{2}\)
Now equation of required line
\[\dfrac{x}{a} + \dfrac{y}{b} = 1\]
\(2{x_1} = a\;,\;2{y_1} = b\)
\(\dfrac{x}{{2{x_1}}} + \dfrac{y}{{2{y_1}}} = 1\)
\(x{y_1} + y{x_1} = 2{x_1}{y_1}\)
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. Midpoint is also known as bisection point which divides the line in two equal parts.
Formula Used:Coordinates of midpoints of line 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 lines.
Equation of line is
\[\dfrac{x}{a} + \dfrac{y}{b} = 1\]
Where
a is x intercept of line
b is y intercept of line
Complete step by step solution:Given: Middle point of line AB is given by
Midpoint=\[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
According to question required line meets meets x-axis and y-axis at the points A and B respectively
Let coordinate of A= \[(a,0)\] and coordinate of B = \[(0,b)\]
Now apply midpoint formula
Midpoint =\[\left( {\dfrac{{\left( {{x_1} + {x_2}} \right)}}{2},\;\dfrac{{\left( {{y_1} + {y_2}} \right)}}{2}} \right)\]
Midpoint=\[\left( {\dfrac{{\left( {a + 0} \right)}}{2},\;\dfrac{{\left( {0 + b} \right)}}{2}} \right)\]
Midpoint of line AB= \[\left( {\dfrac{a}{2},\;\dfrac{b}{2}} \right)\]
middle point of AB be \(\left( {{x_1},{y_1}} \right)\)
\({x_1} = \dfrac{a}{2}\;,\;{y_1} = \dfrac{b}{2}\)
Now equation of required line
\[\dfrac{x}{a} + \dfrac{y}{b} = 1\]
\(2{x_1} = a\;,\;2{y_1} = b\)
\(\dfrac{x}{{2{x_1}}} + \dfrac{y}{{2{y_1}}} = 1\)
\(x{y_1} + y{x_1} = 2{x_1}{y_1}\)
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. Midpoint is also known as bisection point which divides the line in two equal parts.
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