Question

Please help me with geometry question :)

Answer 1

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

$Slope=\dfrac{y_2-y_1}{x_2-x_1}$

$Slope=\dfrac{1-7}{9-1}$

$Slope=\dfrac{-6}{8}$

$Slope=\dfrac{-3}{4}$

Slope of parallel lines are same. So, the slope of parallel line is also $-\dfrac{3}{4}$.

The parallel line passes through D(5,7) with slope $-\dfrac{3}{4}$. So, the equation of the line is

$y-y_1=m(x-x_1)$

$y-7=-\dfrac{3}{4}(x-5)$

$y=-\dfrac{3}{4}(x)+\dfrac{15}{4}+7$

$y=-\dfrac{3}{4}(x)+\dfrac{28+15}{4}$

$y=-\dfrac{3}{4}(x)+\dfrac{43}{4}$              ...(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

$y=-\dfrac{3}{4}(9)+\dfrac{43}{4}$

$y=\dfrac{-27}{4}+\dfrac{43}{4}$

$y=\dfrac{16}{4}$

$y=4$

Therefore, the point at which the intersection of this parallel line be with the line BC is (9,4).