+0  
 
0
1313
2
avatar

Find all points (x,y) that are 13 units away from the point (2,7) and that lie on the line x - 2y = 10

Give your answer as a list of points separated by semicolons, with the points ordered such that their x-coordinates are in increasing order. (So "(1,-3); (2,3); (5,-7)" - without the quotes - is a valid answer format.)

Guest Nov 16, 2014

Best Answer 

 #2
avatar+81077 
+10

This may not be the most direct way......(someone else may nave a better idea)...well, anyway...

We have x - 2y = 10   or   2y = x - 10   or  y = (1/2)x - 5

And a line perpendicular to the given one has a slope of -2

So....using (2, 7) and m = -2, we have

y = 7 = -2(x - 2)

y - 7 = -2x + 4

y = -2x + 11

And setting  -2x + 11 = (1/2)x - 5   we have

-4x + 22 = x - 10   solve for x

5x = 32    →   x = 32/5  = 6.4  and this is the x coordinate of the intersection of both lines

So ...the y intersection point of these two lines is -2(32/5) + 11  =  -64/5 + 11 = -1.8

Now....we need to calculate the distance between this point and (2, 7)....it it's 13, we're through ...(I have a feeling that it isn't though  !!!)

So...d = √[(2- 6.4)^2 + (7 + 1.8)^2] = √96.8   (< 13 !!)...that's good....and you will see why in a second...

Now....we could form two right triangles having their apexes at (2,7), their altitudes = √96.8 and with both of their hypoteneuses = 13. And we could find the length of their legs. The probem is that, we would only know two distances in either direction from (6.4, -1.8), but it wouldn't tell us what those points were. Here's a better soultion:

Imagine a circle with a radius of 13 centered at (2,7)....and we know that such a circle will have to cut the line  x - 2y = 10 in two places..(because the distance from this point to the line < 13)..So....what we're really looking to do is to solve this system:

(x - 2)^2 + (y - 7) = 169

x - 2y   = 10 

So, using substitution, we have x = 2y + 10....and putting this into the equation of the circle, we have

((2y + 10) -2)^2 + (y-7)^2 = 169

(2y+ 8 )^2 + (y-7)^2 = 169     expand this

4y^2 + 32y + 64 + y^2 - 14y + 49 = 169    .... simplify...

5y^2 + 18y + 113 = 169

5y^2 + 18y - 56 = 0      factor this

(5y + 28) (y - 2)  = 0

And setting each factor to 0, we have.......y = -28/5 = -5.6   and y = 2

So x = 2(-5.6) + 10 = -1.2  and x = 2(2) + 10 = 14

So the two points on the line that are 13 units from (2,7) are (-1.2, -5.6) and (14, 2)

Check for yourself that, in fact, both points are 13 units from (2,7)  !!!

Here's a graph of the solution.........https://www.desmos.com/calculator/8gqseydskn

 

CPhill  Nov 16, 2014
Sort: 

2+0 Answers

 #1
avatar+26409 
+5

See Chris's answer below.

An alternative method is to notice that the points are at the intersection of the straight line y = (1/2)x - 5 and a circle of radius 13 centred on (2, 7).

 

Circle equation:  (y - 7)2 + (x - 2)2 = 132

 

Replace y by (1/2)x - 5

 (x/2- 5 - 7)2 + (x - 2)2 = 132

 

Expand the brackets and simplify

5x2/4 - 16x + 148 = 169

 

Subtract 169 from both sides and multiply through the result by 4

5x2 - 64x - 84 = 0

 

This factorises as

(5x + 6)(x - 14) = 0

 

so x = -6/5 (= -1.2) and x = 14

y = -(1/2)(6/5) - 5 = -28/5 = -5.6   and  y = (1/2)14 - 5 = 7 - 5 = 2

 

So the points are (-1.2, -5.6) and (14, 2)

.

Alan  Nov 16, 2014
 #2
avatar+81077 
+10
Best Answer

This may not be the most direct way......(someone else may nave a better idea)...well, anyway...

We have x - 2y = 10   or   2y = x - 10   or  y = (1/2)x - 5

And a line perpendicular to the given one has a slope of -2

So....using (2, 7) and m = -2, we have

y = 7 = -2(x - 2)

y - 7 = -2x + 4

y = -2x + 11

And setting  -2x + 11 = (1/2)x - 5   we have

-4x + 22 = x - 10   solve for x

5x = 32    →   x = 32/5  = 6.4  and this is the x coordinate of the intersection of both lines

So ...the y intersection point of these two lines is -2(32/5) + 11  =  -64/5 + 11 = -1.8

Now....we need to calculate the distance between this point and (2, 7)....it it's 13, we're through ...(I have a feeling that it isn't though  !!!)

So...d = √[(2- 6.4)^2 + (7 + 1.8)^2] = √96.8   (< 13 !!)...that's good....and you will see why in a second...

Now....we could form two right triangles having their apexes at (2,7), their altitudes = √96.8 and with both of their hypoteneuses = 13. And we could find the length of their legs. The probem is that, we would only know two distances in either direction from (6.4, -1.8), but it wouldn't tell us what those points were. Here's a better soultion:

Imagine a circle with a radius of 13 centered at (2,7)....and we know that such a circle will have to cut the line  x - 2y = 10 in two places..(because the distance from this point to the line < 13)..So....what we're really looking to do is to solve this system:

(x - 2)^2 + (y - 7) = 169

x - 2y   = 10 

So, using substitution, we have x = 2y + 10....and putting this into the equation of the circle, we have

((2y + 10) -2)^2 + (y-7)^2 = 169

(2y+ 8 )^2 + (y-7)^2 = 169     expand this

4y^2 + 32y + 64 + y^2 - 14y + 49 = 169    .... simplify...

5y^2 + 18y + 113 = 169

5y^2 + 18y - 56 = 0      factor this

(5y + 28) (y - 2)  = 0

And setting each factor to 0, we have.......y = -28/5 = -5.6   and y = 2

So x = 2(-5.6) + 10 = -1.2  and x = 2(2) + 10 = 14

So the two points on the line that are 13 units from (2,7) are (-1.2, -5.6) and (14, 2)

Check for yourself that, in fact, both points are 13 units from (2,7)  !!!

Here's a graph of the solution.........https://www.desmos.com/calculator/8gqseydskn

 

CPhill  Nov 16, 2014

5 Online Users

avatar
avatar
avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details