Given:
The graph of a triangle ABC.
A line is parallel to AC and passes through the point D.
To find:
The point at which the intersection of this parallel line be with the line BC.
Solution:
From the given figure it is clear that A(1,7), B(9,7), C(9,1) and D(5,7).
Slope of AC is
[tex]Slope=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
[tex]Slope=\dfrac{1-7}{9-1}[/tex]
[tex]Slope=\dfrac{-6}{8}[/tex]
[tex]Slope=\dfrac{-3}{4}[/tex]
Slope of parallel lines are same. So, the slope of parallel line is also [tex]-\dfrac{3}{4}[/tex].
The parallel line passes through D(5,7) with slope [tex]-\dfrac{3}{4}[/tex]. So, the equation of the line is
[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-7=-\dfrac{3}{4}(x-5)[/tex]
[tex]y=-\dfrac{3}{4}(x)+\dfrac{15}{4}+7[/tex]
[tex]y=-\dfrac{3}{4}(x)+\dfrac{28+15}{4}[/tex]
[tex]y=-\dfrac{3}{4}(x)+\dfrac{43}{4}[/tex] ...(i)
From the given graph it is clear that the line BC is a vertical line. So, the x-coordinates of all the points lie on BC are the same, i.e., 9.
Putting x=9, we get
[tex]y=-\dfrac{3}{4}(9)+\dfrac{43}{4}[/tex]
[tex]y=\dfrac{-27}{4}+\dfrac{43}{4}[/tex]
[tex]y=\dfrac{16}{4}[/tex]
[tex]y=4[/tex]
Therefore, the point at which the intersection of this parallel line be with the line BC is (9,4).
