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=(y_2-y_1)/(x_2-x_1)`

`Slope=(1-7)/(9-1)`

`Slope=(-6)/(8)`

`Slope=(-3)/(4)`

Slope of parallel lines are same. So, the slope of parallel line is also
`-(3)/(4)`.

The parallel line passes through D(5,7) with slope
`-(3)/(4)`. So, the equation of the line is

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

`y-7=-(3)/(4)(x-5)`

`y=-(3)/(4)(x)+(15)/(4)+7`

`y=-(3)/(4)(x)+(28+15)/(4)`

`y=-(3)/(4)(x)+(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=-(3)/(4)(9)+(43)/(4)`

`y=(-27)/(4)+(43)/(4)`

`y=(16)/(4)`

`y=4`

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