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1. The lines y=1, y=3, and y=x are drawn along with the point P=(3,0). For certain slopes m, the line P through with slope m will intersect the three given lines so that one of the three points of intersection is the midpoint of the other two points. Find all such values of m.
 

 

2. Determine all values of k for which the points A = (1,2), B= (11,2), and C=(k,6) form the vertices of a right-angled triangle.

 

(This question was posted earlier by Guest, that was me, just posting it again cuz I'm still confused)

(Also, please post an explanation to your solution instead of just answers, thanks)

 Jan 12, 2020
edited by MathCuber  Jan 12, 2020
 #1
avatar+18319 
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Part 1:

One possibility for a line passing through the three given lines so that one of the y-values is the midpoint of the other two is to have the y-values be 1, 3, and 5.

This means that the y-value for the equation y = x is 5.

Since  y  =  x, this means that the x-value must also be 5.

The line must pass through the point (3,0) and also the point (5,5)

Using the two-point form for a straight line:  (y - y1) / (x - x1)  =  (y2 - y1) / (x2 - x1)

Substituting the values:                                 (y - 0) / (x - 3)  =  (5 - 0) / (5 - 3)     --->     y / (x - 3)  =  5 / 2     --->     2y  =  5x - 15

Answer:  5x - 2y  =  15.

 

Since this crosses the line:  y = 1:     5x - 2(1)  =  15     --->     5x  =  17     --->     x  =  3.4

Since this crosses the line:  y =  3:     5x - 2(3)  =  15     --->    5x  =  21     --->     x  =  4.2

The third point is  (5, 5)

You can check that the point  (3, 4.2)  is midway between the points  (1, 3.4)  and  (5 5)

 Jan 12, 2020
 #2
avatar+18319 
0

Part 2:

 

The points A, B, and C will form the vertices of a right triangle is these three points are on the same circle.

Assuming that A and B are the endpoints of a diameter of that triangle:

--  the center of the cirlce will be halfways between the endpoints (1,2) and (11,2) which is the point (6,2)

--  the radius that circle is 5

--  the equation of that circle is  (x - 6)2 + (y - 2)2  =  25

--  for the point (k, 6), the y-value is 6:  (x - 6)2 + (6 - 2)2  =  25     --->     (x - 6)2 + 16  =  25    --->     (x - 6)2  =  9

                  --->     x - 6  =  3     or     x - 6  =  -3     --->     x  =  9     or     x  =  3       

--  if x  =  9:  (9 - 6)2 + (y- 2)2  =  25     --->     9 + (y - 2)2  =  25     --->     (y - 2)2  =  16     --->     y - 2  =  4     or     y - 2  =  -4

                 --->     y  =  6     or     y  =  -2

--->     (9, 6) , (9, -2) , (3, 6) , (3, -2)  

 Jan 12, 2020

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