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###### Difficult coordinate geometry question!

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5 years ago

Points A and B have coords (1, 2) and (5, 5) respectively. The line AB forms the hypotneuse of a right angled, isosceles triangle. Find the two possible coordinates of the other vertex of the triangle.

If someone could work this question through start to finish it would be much appreciated!

If someone could work this question through start to finish it would be much appreciated!

Reply 1

5 years ago

Original post by Tilly M

Points A and B have coords (1, 2) and (5, 5) respectively. The line AB forms the hypotneuse of a right angled, isosceles triangle. Find the two possible coordinates of the other vertex of the triangle.

If someone could work this question through start to finish it would be much appreciated!

If someone could work this question through start to finish it would be much appreciated!

I won't do it from start to finish, but i'll let you start it.

Let the other point be $C$

This means that to get an isoceles triangle, you can either have that the magnitude of $AC=AB$ or $AB=BC$

Use that new information to find out the values.

If the hypotenuse is taken as the diameter of a circle then any point on the surface of said circle connected to the diameter ends will be a right-angle triangle. There are purely algebraic ways to do this but I like this method.

Asking for FULL-SOLUTIONS sounds like you just want someone else to do your homework for you. If you're actually just stuck and need more hints just ask, but providing full solutions is against the rules of the forum. Read the pinned thread for details.

Attachment not found

Asking for FULL-SOLUTIONS sounds like you just want someone else to do your homework for you. If you're actually just stuck and need more hints just ask, but providing full solutions is against the rules of the forum. Read the pinned thread for details.

(edited 5 years ago)

Original post by Ryanzmw

If the hypotenuse is taken as the diameter of a circle then any point on the surface of said circle connected to the diameter ends will be a right-angle triangle. There are purely algebraic ways to do this but I like this method.

Asking for FULL-SOLUTIONS sounds like you just want someone else to do your homework for you. If you're actually just stuck and need more hints just ask, but providing full solutions is against the rules of the forum. Read the pinned thread for details.

Attachment not found

Asking for FULL-SOLUTIONS sounds like you just want someone else to do your homework for you. If you're actually just stuck and need more hints just ask, but providing full solutions is against the rules of the forum. Read the pinned thread for details.

This isn’t my homework it’s just a revision question my teacher made up and I’ve been working on it for a while and just wanted some help until I see my teacher again next week.

Original post by Tilly M

This isn’t my homework it’s just a revision question my teacher made up and I’ve been working on it for a while and just wanted some help until I see my teacher again next week.

Ah fair enough. Hopefully you can appreciate our concern.

Reply 5

5 years ago

Original post by Tilly M

This isn’t my homework it’s just a revision question my teacher made up and I’ve been working on it for a while and just wanted some help until I see my teacher again next week.

Following what Ryanzmw said in post #3, by Thales' Rule (aka "angle in a semicircle is a right angle"), we know that if ACB is a right angle, AB is a diameter of the circle through A, B, and C. Since AB is a diameter, the centre of the circle is the midpoint of AB, which is (3, 7/2). The radius is half the diameter, i.e. half of AB. By Pythagoras' Theorem, AB = sqrt(4^2 + 3^2) = 5, so the radius is 5/2. Now using the coordinate equation for the circle, we get that this circle is (x-3)^2 + (y-7/2)^2 = (5/2)^2. As stated earlier, C lies on this circle, and since the triangle is isosceles, AC = BC, so C is equidistant from A and B, and thus lies on the perpendicular bisector of AB (since the perpendicular bisector is the locus of all points equidistant from two given points). Now, the gradient of AB is (5-2)/(5-1) = 3/4, so the gradient of a line perpendicular to AB is the negative reciprocal of this, i.e. -4/3. Also, as it's a perpendicular bisector, it goes through the midpoint of AB, which we already found to be (3, 7/2). Thus the equation of the perpendicular bisector is y-7/2 = (-4/3)(x-3), and we just need to rearrange to get y=..., then substitute it into the circle equation to find where the line intersects the circle, and this will give the coordinates of the possible positions of C.

Original post by Ryanzmw

Ah fair enough. Hopefully you can appreciate our concern.

Original post by Prasiortle

Following what Ryanzmw said in post #3, by Thales' Rule (aka "angle in a semicircle is a right angle"), we know that if ACB is a right angle, AB is a diameter of the circle through A, B, and C. Since AB is a diameter, the centre of the circle is the midpoint of AB, which is (3, 7/2). The radius is half the diameter, i.e. half of AB. By Pythagoras' Theorem, AB = sqrt(4^2 + 3^2) = 5, so the radius is 5/2. Now using the coordinate equation for the circle, we get that this circle is (x-3)^2 + (y-7/2)^2 = (5/2)^2. As stated earlier, C lies on this circle, and since the triangle is isosceles, AC = BC, so C is equidistant from A and B, and thus lies on the perpendicular bisector of AB (since the perpendicular bisector is the locus of all points equidistant from two given points). Now, the gradient of AB is (5-2)/(5-1) = 3/4, so the gradient of a line perpendicular to AB is the negative reciprocal of this, i.e. -4/3. Also, as it's a perpendicular bisector, it goes through the midpoint of AB, which we already found to be (3, 7/2). Thus the equation of the perpendicular bisector is y-7/2 = (-4/3)(x-3), and we just need to rearrange to get y=..., then substitute it into the circle equation to find where the line intersects the circle, and this will give the coordinates of the possible positions of C.

Thank you both for all your help I understand it now

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