
In an isosceles triangle \[ABC\] the coordinates of the points B and C on the base \[BC\] are respectively \[(1,2)\] and \[(2,1)\] if the equation of the line \[AB\] is \[y = 2x\] then the equation of the line \[AC\] is
A. \[y = \dfrac{1}{2}(x - 1)\]
B. \[y = \dfrac{x}{2}\]
C. \[y = x - 1\]
D. \[2y = x + 3\]
Answer
167.7k+ views
Hint: To begin, consider the property that its altitude is its median. Then, using the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\] determine the altitude equation, where m is the slope and (x1,y1) is the mid-point of the base. Then, given the fact that AB and altitude overlap at A, locate the coordinates of A. Find the equation of a line starting at point A.
Formula Used: The two line’s slope intersecting at point is,
\[{{\rm{m}}_1} \cdot {{\rm{m}}_2} = - 1\]
Complete step by step solution: We have been provided in the question that,
In an isosceles triangle \[ABC\] the coordinates of the points B and C on the base \[BC\] are respectively \[(1,2)\] and \[(2,1)\]
The perpendicular bisector of the base of an isosceles triangle will always pass through the third vector, as we know.
Let us have the midpoint of \[{\rm{BC}}\] as,
\[ = \left[ {\dfrac{{\dfrac{1}{2} + 1}}{{1 + \dfrac{1}{2}}}} \right]\]
On simplifying the numerator and denominator, we get
\[ = - \dfrac{1}{3}\]
We already know that slope can be found using
\[{{\rm{m}}_1} = \dfrac{{{{\rm{y}}_2} - {{\rm{y}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}}\]
On simplification, we get
\[ = \dfrac{1}{2}\]
We have already know that the two line’s slope intersecting at point is,
\[{{\rm{m}}_1} \cdot {{\rm{m}}_2} = - 1\]
As, we know the value of \[{{\rm{m}}_1}\] and substitute the value and find \[{{\rm{m}}_2}\] we get
\[\therefore {{\rm{m}}_2} = 1\]
Hence, equation becomes
The line’s equation with slope and a point is,
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Now, we have to substitute all the obtained values before, we get
\[y - 1 = \dfrac{1}{2}\left( {x - 2} \right)\]
On simplification, we get
\[ \Rightarrow y = \dfrac{x}{2}\]
Therefore, the equation of the line \[AC\] is \[y = \dfrac{x}{2}\].
Option ‘B’ is correct
Note: Students should be cautious while solving for values of x and y from the equations to obtain the co-ordinates, we will determine their slope and so find the equation; otherwise, the solution may be incorrect.
Formula Used: The two line’s slope intersecting at point is,
\[{{\rm{m}}_1} \cdot {{\rm{m}}_2} = - 1\]
Complete step by step solution: We have been provided in the question that,
In an isosceles triangle \[ABC\] the coordinates of the points B and C on the base \[BC\] are respectively \[(1,2)\] and \[(2,1)\]
The perpendicular bisector of the base of an isosceles triangle will always pass through the third vector, as we know.
Let us have the midpoint of \[{\rm{BC}}\] as,
\[ = \left[ {\dfrac{{\dfrac{1}{2} + 1}}{{1 + \dfrac{1}{2}}}} \right]\]
On simplifying the numerator and denominator, we get
\[ = - \dfrac{1}{3}\]
We already know that slope can be found using
\[{{\rm{m}}_1} = \dfrac{{{{\rm{y}}_2} - {{\rm{y}}_1}}}{{{{\rm{x}}_2} - {{\rm{x}}_1}}}\]
On simplification, we get
\[ = \dfrac{1}{2}\]
We have already know that the two line’s slope intersecting at point is,
\[{{\rm{m}}_1} \cdot {{\rm{m}}_2} = - 1\]
As, we know the value of \[{{\rm{m}}_1}\] and substitute the value and find \[{{\rm{m}}_2}\] we get
\[\therefore {{\rm{m}}_2} = 1\]
Hence, equation becomes
The line’s equation with slope and a point is,
\[y - {y_1} = m\left( {x - {x_1}} \right)\]
Now, we have to substitute all the obtained values before, we get
\[y - 1 = \dfrac{1}{2}\left( {x - 2} \right)\]
On simplification, we get
\[ \Rightarrow y = \dfrac{x}{2}\]
Therefore, the equation of the line \[AC\] is \[y = \dfrac{x}{2}\].
Option ‘B’ is correct
Note: Students should be cautious while solving for values of x and y from the equations to obtain the co-ordinates, we will determine their slope and so find the equation; otherwise, the solution may be incorrect.
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