Question

1. The ordinate of a point which is √29 units from (5,1) is twice its abscissa. Find the point.
2. A point is at a distance 4√2 from (-3/2,-5/2) and at a distance 2√5 from (9/2,-5/2). Find the point.
3. Find the coordinTES OF A POINT EQUIDISTANT FROM (-2,8), (1,-1) and (-8, -4).

thank you!

Answer 1

Task 1. The ordinate of a point which is $\sqrt{29}$ units from (5,1) is twice its abscissa. Find the point.

Solution. Let $A=(x,y)$ be the point which we are looking for, and $B=(5,1)$. Then the ordinate $y$ of $A$ is twice its abscissa $x$, that is $y=2x$, and so $A=(x,2x)$.

Moreover, the distance $AB=\sqrt{29}$.

We should find $x$.

Let us write the distance $AB$:

$AB=\sqrt{(x-5)^{2}+(2x-1)^{2}}=\sqrt{29},$

whence

$(x-5)^{2}+(2x-1)^{2}=29,$

$x^{2}-10x+25+4x^{2}-4x+1=29$

$5x^{2}-14x+26-29=0$

$5x^{2}-14x-3=0,$

$D=14^{2}+4*5*3=256=16^{2}$

$x_{1}=\frac{14+16}{2*5}=\frac{30}{10}=3,\qquad x_{2}=\frac{14-16}{2*5}=-\frac{2}{10}=-0.3.$

Hence

$y_{1}=2x_{1}=2*3=6,\qquad y_{2}=2x_{2}=2*(-0.2)=-0.4.$

Answer. There are two points

$A_{1}(3,6),\qquad A_{2}(-0.2,-0.4).$

.

Task 2. A point $A(x,y)$ is at a distance $4\sqrt{2}$ from $B=(-3/2,-5/2)$ and at a distance $2\sqrt{5}$ from $C=(9/2,-5/2)$. Find the point.

Solution. Let us write the formulas for the distances $AB$ ad $AC$:

$AB=\sqrt{(x-(-3/2))^{2}+(y-(-5/2))^{2}}=4\sqrt{2}$

$AC=\sqrt{(x-9/2)^{2}+(y-(-5/2))^{2}}=2\sqrt{5}$

From the first equation we get

$(x+1.5)^{2}+(y+2.5)^{2}=(4\sqrt{2})^{2}$

$x^{2}+3x+2.25+y^{2}+5y+6.25=32$

$x^{2}+3x+y^{2}+5y+8.5=32$

$x^{2}+3x+y^{2}+5y=23.5.$

The second equation gives

$(x-4.5)^{2}+(y+2.5)^{2}=(2\sqrt{5})^{2}$

$x^{2}-9x+20.25+y^{2}+5y+6.25=50$

$x^{2}-9x+y^{2}+5y+26.5=50$

$x^{2}-9x+y^{2}+5y=23.5.$

Thus we get the following system

\[ \begin{cases}x^{2}+3x+y^{2}+5y=23.5\\

x^{2}-9x+y^{2}+5y=23.5\end{cases} \]

Subtracting the second equation from the first we get

$x^{2}+3x+y^{2}+5y-(x^{2}-9x+y^{2}+5y)=23.5-23.5$

$12x=0\qquad\Rightarrow\qquad x=0.$

Sinstituting $x$ into the first equation we obtain

$y^{2}+5y=23.5$

$y^{2}+5y-23.5=0$

$D=25+4*23.5=119$

$y_{1}=\frac{-5+\sqrt{119}}{2},\qquad y_{2}=\frac{-5-\sqrt{119}}{2}.$

Answer. There are two points:

$A_{1}=\left(0,\frac{-5+\sqrt{119}}{2}\right),\qquad A_{2}=\left(0,\frac{-5-\sqrt{119}}{2}\right).$

Task 3. Find the coordintes of a point $P=(x,y)$ equidistant from $A=(-2,8)$, $B=(1,-1)$ and $C=(-8,-4)$.

Solution. Let us write squares of distances $PA$, $PB$ and $PC$:

$PA^{2}=(x+2)^{2}+(y-8)^{2}=x^{2}+4x+4+y^{2}-16y+64=x^{2}+4x+y^{2}-16y+68,$

$PB^{2}=(x-1)^{2}+(y+1)^{2}=x^{2}-2x+1+y^{2}+2y+1=x^{2}-2x+y^{2}+2y+2,$

$PC^{2}=(x+8)^{2}+(y+4)^{2}=x^{2}+16x+64+y^{2}+8y+16=x^{2}+16x+y^{2}+8y+80.$

We have that these distances coincides, so

$PA^{2}=PB^{2},\qquad PA^{2}=PC^{2}.$

Let us write down $PA^{2}=PB^{2}$:

$x^{2}+4x+y^{2}-16y+68=x^{2}-2x+y^{2}+2y+2,$

$4x-16y+68=-2x+2y+2,$

$6x-18y=-66,$

$x-3y=-11.$

Similarly, write down $PA^{2}=PC^{2}$:

$x^{2}+4x+y^{2}-16y+68=x^{2}+16x+y^{2}+8y+80$

$4x-16y+68=16x+8y+80$

$-12x-24y=12$

$x+2y=-1.$

Thus we get the following system

\[ \begin{cases}x-3y=-11\\

x+2y=-1.\end{cases} \]

Subtracting the second equation from the first we get

$x-3y-(x+2y)=-11-(-1)$

$-5y=-10\qquad\Rightarrow\qquad y=2$

Hence, from $x+2y=-1$ we get

$x=-1-2y=-1-2*2=-5.$

Answer. $P(-5,2).$