
The points A(5,4), B(-3,-2) and C(1,-8) are the vertices of a triangle ABC. Find the equations of median AD and line parallel to AC passing through point B.
Answer
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Hint: First of all, try to represent the given conditions of the question in the form of a figure. Then with the help of the figure, evaluate the required coordinates, the equations of the line and the other information by using the given values.
Complete step-by-step answer:
The points A(5,4), B(-3,-2) and C(1,-8) are the vertices of a triangle ABC with the median AD, this means that it is the mid-point of the side BC.
So, the coordinates of the mid-point D can be taken as
$D = \left( {\dfrac{{ - 3 + 1}}{2},\dfrac{{ - 2 - 8}}{2}} \right)$
$ \Rightarrow D = \left( {\dfrac{{ - 2}}{2},\dfrac{{ - 10}}{2}} \right)$
$\therefore D = \left ({ - 1, - 5} \right)$
Now, the equation of the line is given as
$y - {y_1} = \dfrac{{{y_2} - {y_1}}} {{{x_2} - {x_1}}} (x - {x_1})$
Therefore, the equation of the line AD can be written as
$y - (- 5) = \dfrac{{4 - ( - 5)}}{{5 - ( - 1)}}(x - ( - 1))$
$ \Rightarrow y + 5 = \dfrac{{4 + 5}}{{5 + 1}}(x + 1)$
$ \Rightarrow y + 5 = \dfrac{9}{6}(x + 1)$
$ \Rightarrow y + 5 = \dfrac{3}{2}(x + 1)$
$ \Rightarrow 2(y + 5) = 3(x + 1) $
$ \Rightarrow 2y + 10 = 3x + 3$
$\therefore 3x - 2y = 7$
The slope of the equation is given as
$m = \dfrac{{{y_2} - {y_1}}} {{{x_2} - {x_1}}} $
Therefore, the slope of AC can be written as
${m_{AC}} = \dfrac{{ - 8 - 4}}{{1 - 5}}$
$ \Rightarrow {m_{AC}} = \dfrac{{ - 12}}{{ - 4}}$
$\therefore {m_{AC}} = 3$
Now, the equation of the line parallel to AC passing through the point B
$y - ( - 2) = 3(x - (3))$
$ \Rightarrow y + 2 = 3x + 9$
$\therefore 3x - y = - 7$
Hence, the equation for the median AD is $3x - 2y = 7$and the equation for the line parallel to AC passing through the point B is $3x - y = - 7$.
Note: The students can be mistaken in finding the equations of the line as the various formulas are used to reach this equation, like slope is needed to be calculated first, sometimes the coordinates are also calculated. So, the students need an adequate knowledge of these concepts before proceeding for these solutions.
Complete step-by-step answer:
The points A(5,4), B(-3,-2) and C(1,-8) are the vertices of a triangle ABC with the median AD, this means that it is the mid-point of the side BC.
So, the coordinates of the mid-point D can be taken as
$D = \left( {\dfrac{{ - 3 + 1}}{2},\dfrac{{ - 2 - 8}}{2}} \right)$
$ \Rightarrow D = \left( {\dfrac{{ - 2}}{2},\dfrac{{ - 10}}{2}} \right)$
$\therefore D = \left ({ - 1, - 5} \right)$
Now, the equation of the line is given as
$y - {y_1} = \dfrac{{{y_2} - {y_1}}} {{{x_2} - {x_1}}} (x - {x_1})$
Therefore, the equation of the line AD can be written as
$y - (- 5) = \dfrac{{4 - ( - 5)}}{{5 - ( - 1)}}(x - ( - 1))$
$ \Rightarrow y + 5 = \dfrac{{4 + 5}}{{5 + 1}}(x + 1)$
$ \Rightarrow y + 5 = \dfrac{9}{6}(x + 1)$
$ \Rightarrow y + 5 = \dfrac{3}{2}(x + 1)$
$ \Rightarrow 2(y + 5) = 3(x + 1) $
$ \Rightarrow 2y + 10 = 3x + 3$
$\therefore 3x - 2y = 7$
The slope of the equation is given as
$m = \dfrac{{{y_2} - {y_1}}} {{{x_2} - {x_1}}} $
Therefore, the slope of AC can be written as
${m_{AC}} = \dfrac{{ - 8 - 4}}{{1 - 5}}$
$ \Rightarrow {m_{AC}} = \dfrac{{ - 12}}{{ - 4}}$
$\therefore {m_{AC}} = 3$
Now, the equation of the line parallel to AC passing through the point B
$y - ( - 2) = 3(x - (3))$
$ \Rightarrow y + 2 = 3x + 9$
$\therefore 3x - y = - 7$
Hence, the equation for the median AD is $3x - 2y = 7$and the equation for the line parallel to AC passing through the point B is $3x - y = - 7$.
Note: The students can be mistaken in finding the equations of the line as the various formulas are used to reach this equation, like slope is needed to be calculated first, sometimes the coordinates are also calculated. So, the students need an adequate knowledge of these concepts before proceeding for these solutions.
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