Question

Find the possible values forif the points with the given coordinates are the indicated distance apart.

$\left(1,a\right)$, $\left(-3,2\right)$;$d=5$

Short answer

The possible values for a are5 and $\mathbf{-}\mathbf{1}$.

Step-by-step solution

Step 1. Given. 

Given data is$\left(1,a\right),\left(-3,2\right);d=5\left(1,a\right),\left(-3,2\right);d=5$

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:

width="191">d=x2−x12+y2−y12

Step 3. Find the possible values for a.

It is given that the two points are $\left(1,a\right)$and $\left(-3,2\right)$. Therefore, the values of $x_1$, $y_1$, $x_2$and$y_2$ are 1, a,$-3$ and 2 respectively.

It is also given that the distance between the given points is 5. Therefore, the value d is 5.

Therefore, it is obtained that:

$$\begin{gathered} d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2} \\ 5=\sqrt{{\left(-3-1\right)}^2+{\left(2-a\right)}^2} \\ 5=\sqrt{{\left(-4\right)}^2+{\left(2-a\right)}^2} \\ 5=\sqrt{16+2^2-2\left(2\right)\left(a\right)+a^2}\left(\because {\left(a-b\right)}^2=a^2-2ab+b^2\right) \\ 5=\sqrt{a^2-4a+20} \\ {\left(5\right)}^2={\left(\sqrt{a^2-4a+20}\right)}^2 \\ 25=a^2-4a+20 \\ 0=a^2-4a+20-25 \\ 0=a^2-4a-5 \\ 0=a^2-5a+a-5 \\ 0=a\left(a-5\right)+1\left(a-5\right) \\ 0=\left(a-5\right)\left(a+1\right) \\ a-5=0\text{or}a+1=0 \\ a=5\text{or}a=-1 \end{gathered}$$

Therefore, the possible values for a are 5 and $-1$.