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# Find all points that are 13 units away from the point and that lie on the line .

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Find all points  that are 13 units away from the point  and that lie on the line

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

Nov 16, 2014

#2
+94619
+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

Nov 16, 2014

#1
+27396
+5

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)

.

Nov 16, 2014
#2
+94619
+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