Answer to Question #278878 in Analytic Geometry for Biniyam

Question

1 Determine k for which line (k + 2)x + (k^2 -9)y + 3k^2 - 8k + 5 =0 is 1, parallel to y axis 2, parallel to x axis 3, pass through origin 4, pass through (1,1)

Answer 1

1.

parallel to $x=0$

then:

$k^2-9=0$

$k=\pm 3$

2.

parallel to $y=0$

then:

$k+2=0$

$k=-2$

3.

$3k^2 - 8k + 5 =0$

$k=\frac{8\pm \sqrt{64-60}}{6}$

$k_1=1,k_2=5/3$

4.

$(k + 2) + (k^2 -9) + 3k^2 - 8k + 5 =0$

$4k^2-7k-2=0$

$k=\frac{7\pm \sqrt{49+32}}{8}$

$k_1=-1/4,k_2=2$