Q9. - Algebra 2

Question

Which of the following point is closest to $\left(2,-4\right)$.

$\begin{array}{l}\left(A\right)\left(3,1\right) \left(B\right)\left(-2,0\right)\\ \left(C\right)\left(1,5\right) \left(D\right)\left(4,-2\right)\end{array}$

Step-by-Step Solution

Answer

The distance of point $P\left(2,-4\right)$ from the point $D\left(4,-2\right)$ is least. Hence the point $P\left(2,-4\right)$ is closest to $D\left(4,-2\right)$. So, option (D) is correct.

Step 1. Write down the given information.

The given points are $A\left(3,1\right), B\left(-2,0\right), C\left(1,5\right)\text{ and }D\left(4,-2\right)$ whose distance from the fixed point say $P\left(2,-4\right)$ has to determine.

Step 2. Calculation.

The distance $\left(d\right)$ from the point $P\left(2,-4\right)$ is evaluated from the points $A\left(3,1\right), B\left(-2,0\right), C\left(1,5\right)\text{ and }D\left(4,-2\right)$.

$\begin{array}{c}\left|d_{PA}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(3-2\right)}^2+{\left(1+4\right)}^2}\\ =\sqrt{26}\text{ units}\end{array}$

$\begin{array}{c}\left|d_{PB}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-2-2\right)}^2+{\left(0+4\right)}^2}\\ =4\sqrt{2}\text{ units}\end{array}$

$\begin{array}{c}\left|d_{PC}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(1-2\right)}^2+{\left(5+4\right)}^2}\\ =\sqrt{82}\text{ units}\end{array}$

$\begin{array}{c}\left|d_{PD}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(4-2\right)}^2+{\left(-2+4\right)}^2}\\ =2\sqrt{2}\text{ units}\end{array}$

Step 3. Conclusion.

The distance of point $P\left(2,-4\right)$ from the point $D\left(4,-2\right)$ is least. Hence the point $P\left(2,-4\right)$ is closest to $D\left(4,-2\right)$. So, option (D) is correct.