Q57. - Algebra 1

Question

Find the possible values for a if the points with the given coordinates are the indicated distance apart.

$\left(5,-2\right)$, $\left(a,-3\right)$; $d=\sqrt{170}$

Step-by-Step Solution

Answer

The possible values for a are 18 and $\mathbf{-}\mathbf{8}$.

Step 1. Given.

Coordinates are $\left(5,-2\right),\left(a,-3\right);d=\sqrt{170}$

Step 2. Write the distance formula to find the distance between the given two points.

The distance (d) between the points $\left(x_1,y_1\right)$ and $\left(x_2,y_2\right)$ is given by:

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Step 3. Find the possible values for a .

It is given that the two points are $\left(5,-2\right)$ and $\left(a,-3\right)$. Therefore, the values of $x_1$, ,$y_1$ $x_2$ and $y_2$ are 5, $-2$, a and $-3$ respectively.

It is also given that the distance between the given points is $\sqrt{170}$. Therefore, the value of d is $\sqrt{170}$.

Therefore, it is obtained that:

$$\begin{gathered} d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ \sqrt{170}=\sqrt{{\left(a-5\right)}^2+{\left(-3-\left(-2\right)\right)}^2} \\ \sqrt{170}=\sqrt{{\left(a-5\right)}^2+{\left(-3+2\right)}^2} \\ \sqrt{170}=\sqrt{a^2-2\left(a\right)\left(5\right)+5^2+{\left(-1\right)}^2}\left(\because {\left(a-b\right)}^2=a^2-2ab+b^2\right) \\ \sqrt{170}=\sqrt{a^2-10a+26} \\ {\left(\sqrt{170}\right)}^2={\left(\sqrt{a^2-10a+26}\right)}^2 \\ 170=a^2-10a+26 \\ 0=a^2-10a+26-170 \\ 0=a^2-10a-144 \\ 0=a^2-18a+8a-144 \\ 0=a\left(a-18\right)+8\left(a-18\right) \\ 0=\left(a-18\right)\left(a+8\right) \\ a-18=0\text{ or }a+8=0 \\ a=18\text{ or }a=-8 \end{gathered}$$

Therefore, the possible values for a are 18 and $-8$.