A point $P(x,y)$ has a distance $5\sqrt{2}$ units from $Q(4, -7)$ and a distance $\sqrt{106}$ units from $R(-6,5)$. Knowing that, find $P$.
the image is exactly the set of problem that our professor given to us
I assume that the 3 problems are somewhat related, and the only formula that she taught us is the distance formula, which is $D =\sqrt{(x_2−x)^2+(y_2−y)^2}$
On
Notice, distance between $P(x, y)$ & $Q(4, -7)$ is $5\sqrt 2$ hence, $$\sqrt{(x-4)^2+(y-(-7))^2}=5\sqrt 2$$ $$x^2+y^2-8x+14y+15=0\tag 1$$ distance between $P(x, y)$ & $R(-6, 5)$ is $\sqrt {106}$ hence, $$\sqrt{(x-(-6))^2+(y-5)^2}=\sqrt {106}$$ $$x^2+y^2+12x-10y-45=0\tag 2$$ subtracting (1) from (2), one should get $$20x-24y-60=0$$$$\implies y=\frac{5x-15}{6}\tag 3$$ setting the value of $y$ in (1), one should get $$x^2+\left(\frac{5x-15}{6}\right)^2-8x+14\left(\frac{5x-15}{6}\right)+15=0$$ $$61x^2+18x-720=0$$ $$x=\frac{-9\pm \sqrt{44001}}{61}$$ now, set the values of $x$ in (3), find corresponding values of $y$
Note: There are two points $P(x, y) $ lying either side of the line $QR$ joining the points $Q(4, 7)$ & $R(-6, 5)$.
$$\sqrt{(x_p-4)^2+(y_p+7)^2}=5\sqrt{2}$$ $$\sqrt{(x_p+6)^2+(y_p-5)^2}=\sqrt{106}$$
Here, we do:
$${(x_p-4)^2+(y_p+7)^2}=50$$ $${(x_p+6)^2+(y_p-5)^2}=106$$
Now we, expand above two formulas (for simplicity, i use x and y):
$$x^2 -8x+16+y^2+14y+49=50$$ $$x^2+12x+36+y^2-10y+25=106$$
Some modification, $$x^2+y^2-8x+14y=-15 \ \ \ \ (I)$$ $$x^2+y^2+12x-10y=40 \ \ \ \ (II)$$
here, we multiplie them by $-1$ then summation, we get:
$$20x+-24y=65$$
therfore, we have:
$$x=\frac{65+24y}{20} \ \ \ \ \ (**)$$
by putting (**) in (I) or (II), you can calculate the possible answers