# Question 36cf7

Jan 18, 2018

#### Answer:

a.) : P(1,1).

b.) : Contd.........

#### Explanation:

Part a.): Suppose that, the required point (reqd. pt.) $P = P \left(x , y\right)$.

The given pts. are $A \left(2 , - 1\right) , B \left(- 1 , 5\right) \mathmr{and} C \left(2 , 5\right)$.

$\therefore \vec{P A} = \left(2 , - 1\right) - \left(x , y\right) = \left(2 - x , - 1 - y\right)$.

$\vec{P B} = \left(- 1 - x , 5 - y\right) , \mathmr{and} \vec{P C} = \left(2 - x , 5 - y\right)$.

Recall that, $\left({x}_{1} , {y}_{1}\right) + \left({x}_{2} , {y}_{2}\right) = \left({x}_{1} + {x}_{2} , {y}_{1} + {y}_{2}\right)$.

$\therefore \vec{P A} + \vec{P B} + \vec{P C} ,$

$\left(2 - x , - 1 - y\right) + \left(- 1 - x , 5 - y\right) + \left(2 - x , 5 - y\right)$

$= \left(2 - x - 1 - x + 2 - x , - 1 - y + 5 - y + 5 - y\right) ,$

$= \left(3 - 3 x , 9 - 3 y\right)$.

$\text{But, } \vec{P A} + \vec{P B} + \vec{P C} = 0 = \left(0 , 0\right)$.

$\therefore \left(3 - 3 x , 9 - 3 y\right) = \left(0 , 0\right)$.

Knowing that, $\left({x}_{1} , {y}_{1}\right) = \left({x}_{2} , {y}_{2}\right) \Rightarrow {x}_{1} = {x}_{2} , {y}_{1} = {y}_{2}$, we get,

$\left(3 - 3 x , 9 - 3 y\right) = \left(0 , 0\right) \Rightarrow 3 - 3 x = 0 , 9 - 9 y = 0 , \mathmr{and} ,$

$x = 1 , y = 1 , \text{ which give, } P = P \left(x , y\right) = P \left(1 , 1\right)$.

Jan 18, 2018

#### Answer:

Point P has the coordinates $\left(1 , 3\right)$.

The points $\left(- 8 , 0\right)$ and $\left(0 , 0\right)$ will each make a different scalene triangle, but no solution is available that will make an equilateral triangle.

#### Explanation:

For any of these it makes sense to draw a diagram if you can. I haven't here, but you should. Do as I say, not as I do. ;-)

For part (a), we can treat each direction separately. We want the sum in $x$ to be zero and the sum in $y$ to be zero. If we decide that point P has the coordinates $\left(x , y\right)$, then we can say that in the $x$ dimension:

$\left(x - 2\right) + \left(x - \left(- 1\right)\right) + \left(x - 2\right) = 0$

I've just subtracted each of the $x$ values of the known points from $x$, that is, found their distances from $x$ in the x-direction.

$\left(x - 2\right) + \left(x + 1\right) + \left(x - 2\right) = 0$

$3 x - 3 = 0$

$x - 1 = 0$

$x = 1$

Same method for $y$:

$\left(y - \left(- 1\right)\right) + \left(y - 5\right) + \left(y - 5\right) = 0$

$3 y - 9 = 0$

$y = 3$

So $P$ has the coordinates $\left(1 , 3\right)$. Add it to your diagram of the other three points and see whether it looks like it's in the centre.

For part (b) we have the coordinates of two points on a triangle, $\left(0 , 6\right)$ and $\left(- 4 , 3\right)$. We also know that, since the third point sits on the $x$ axis, it's $y$ coordinate will be $0$, so the third point is $\left(x , 0\right)$.

Draw a diagram!

To make the triangle scalene, we would need two of its sides to be the same length. To make it equilateral, all three sides would need to be the same length.

Let's calculate the length of the side we already know:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$= \sqrt{{\left(0 - \left(- 4\right)\right)}^{2} + {\left(3 - 6\right)}^{2}} = \sqrt{16 + 9} = 5$ units

We need to choose values of $x$ such that the distances from the point (x,0) to the point $\left(0 , 6\right)$ and/or from the point $\left(x , 0\right)$ to the point $\left(- 4 , 3\right)$ are equal to 5. If only one of them is, we will have a scalene triangle, if both of them are we will have an equilateral triangle.

Try first for $\left(0 , 6\right)$:

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$5 = \sqrt{{\left(0 - x\right)}^{2} + {\left(6 - 0\right)}^{2}}$

$5 = \sqrt{36 + {x}^{2}}$

${5}^{2} = 25 = \left(36 + {x}^{2}\right)$

${x}^{2} = - 11$

Now, in the real numbers, we cannot do this (in imaginary numbers we can, but you're probably not studying that at the moment).

Since we can't find a value that makes this side have a length of $5$, we definitely can't make it an equilateral triangle.

Let's try for $\left(- 4 , 3\right)$, see if we can't at least salvage a scalene triangle.

$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

$5 = \sqrt{{\left(- 4 - x\right)}^{2} + {\left(3 - 0\right)}^{2}}$

$5 = \sqrt{{\left(- 4 - x\right)}^{2} + 9}$

${\left(- 4 - x\right)}^{2} = 16 + 4 x + 4 x + {x}^{2} = 16 + 8 x + {x}^{2}$

$5 = \sqrt{16 + 8 x + {x}^{2} + 9} = \sqrt{{x}^{2} + 8 x + 25}$

(I'm gonna go around this a long way to make sure it's clear:

${5}^{2} = {x}^{2} + 8 x + 25$

${x}^{2} + 8 x = 0$

I would use the quadratic formula to solve this quadratic: use your favourite method.

$x = \frac{- 8 \pm \sqrt{{8}^{2} - 0}}{2} = - 8 \mathmr{and} 0$

So there are two solutions - two points on the $x$ axis - that will give us a scalene triangle, but no solutions that will give us an equilateral triangle.

Test this on your diagram.