Answer to Question #241042 in Analytic Geometry for Lady viper

Question

The abiscissa of a point in the 4th quadrant is numerically three times its ordinate and is 10 units from (-2,4). Find the point.

Answer 1

 Let the required point be $(x,y).$  Then

$x<0, y<0, x=3y$

$d^2= (x-(-2)^2+(y-4)^2=10^2$

$$(3y+2)^2+(y-4)^2=100$$

$$9y^2+12y+4+y^2-8y+16-100=0$$

$$10y^2+4y-80=0$$

$$5y^2+2y-40=0$$

$$D=(2)^2-4(5)(-40)=804$$

$$y=\dfrac{-2\pm\sqrt{804}}{2(5)}=\dfrac{-1\pm\sqrt{201}}{5}$$

Since $y<0,$ we take $y=\dfrac{-1-\sqrt{201}}{5}.$

$x=3(\dfrac{-1-\sqrt{201}}{5})$

Point $(-\dfrac{3+3\sqrt{201}}{5}, -\dfrac{1+\sqrt{201}}{5})$