Question

The distance between Ql-1, a) and P(3. – 2) is 4/5. Find all possible values of a.

Answer 1

Q (-1, a)

P (3, -2)

distance's formula:

$\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}\text{ }$

So:

$d\text{ = }\sqrt[]{(3+1)^2+(-2-a)^2}\text{ = }4\sqrt[]{5}$

$\sqrt[]{16+4+4a+a^2\text{ }}=\text{ 4}\sqrt[]{5}$$\sqrt[]{20+4a+a^2\text{ }}=\text{ 4}\sqrt[]{5}$$20+4a+a^2\text{ = 16 x 5}$$\begin{gathered} 20+4a+a^2\text{ = 80} \\ 20+4a+a^2-\text{ 80 = 0 } \\ -60+4a+a^2\text{ = 0} \\ (a\text{ + 10 )(a - 6) = 0} \\ a\text{ = 6 or a = -10} \end{gathered}$