Question

14. Line m and triangle PQR are shown on the graph below

Greg is creating triangle JKL to be s
is parallel to one side of triangle PQR. Select all the points that co
of point L.
D. (2,3)
A. (-2, 0)
eating triangle JKL to be similar to triangle POR. Each side of triangle JRL
Hect all the points that could be the location
E. (2.5, 3.5)
B. (-1,-1.5)
F. (5.5, 8)
C. (0,0)

Answer 1

Sure, I'd be happy to help!

To find the points that could be the location of point L, we need to find the points that are on the line m and parallel to one side of triangle PQR.

From the given information, we know that line m is the line that passes through point D (2,3) and point L. We also know that triangle PQR is a right triangle, so one of its sides is the hypotenuse of a right triangle.

Let's call the length of the hypotenuse of triangle PQR "x". Then, we can write the equation of the line m as:

y = mx + 2

Now, we need to find the points that are on this line and parallel to one side of triangle PQR. To do this, we can use the fact that the sum of the distances from a point to two lines is equal to the distance from the point to the intersection of the lines.

Let's call the point where the line m intersects the side of triangle PQR "P". Then, we can write the equation of the line PQ as:

y = qx + p

where p is the y-coordinate of point P.

Now, we can find the points that are on the line m and parallel to the line PQ by using the fact that the sum of the distances from a point to two lines is equal to the distance from the point to the intersection of the lines. Specifically, we can use the fact that:

distance from point L to line m = distance from point L to line PQ

We can write this as:

|LM| = |LP|

where |LM| is the distance from point L to line m, and |LP| is the distance from point L to line PQ.

Now, we can substitute the equations of the lines m and PQ into this equation to get:

|LM| = |2x + 2| = |qx + p|

Simplifying this equation, we get:

|LM| = qx + 2

Now, we can find the points that satisfy this equation by solving for x. Specifically, we can use the fact that the equation of the line m is y = mx + 2, so we can substitute y = qx + 2 into this equation to get:

qx + 2 = mx + 2

Simplifying this equation, we get:

qx = mx

Now, we can solve for x to find the points that are on the line m and parallel to one side of triangle PQR. Specifically, we can use the fact that the slope of the line m is m, so we can write:

mx = 0

Solving for x, we get:

x = 0

Therefore, the points that could be the location of point L are (0,2) and (0,3).

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The points that could be the location of point L are (0,2) and (0,3).

Explanation: