Chapter 1, Problem 8MCQ

To determine

To find the coordinates of the midpoint and then distance between each pair of points

Answer to Problem 8MCQ

Coordinates L $\left(-\frac{5}{2},\frac{1}{2}\right)$

Distance between

JK = 7.07

JL = 3.53

KL = 3.53

Explanation of Solution

Given information:

Coordinates J (-2, 4) and K (-3,-3)

Formula used:

To calculate coordinate

  $x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$

To calculate the distance

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

Calculation: calculating the coordinates

Let the coordinates of J (-2, 4) be $x_1$ , $y_1$ and K (-3, -3) be $x_2,y_2$

Substituting values in the formula

  $x=\frac{-2-3}{2},y=\frac{4-3}{2}$

Solving it further

  $x=-\frac{5}{2},y=\frac{1}{2}$

Therefore the coordinates of the midpoint are

Coordinates of midpoint L $\left(-\frac{5}{2},\frac{1}{2}\right)$

Now calculating distance between each of the points

First take J (-2, 4) and K (-3, -3)

Substituting values in the formula

Distance = $\sqrt{{\left(-3+2\right)}^2+{\left(-3-4\right)}^2}$

Solving further

Distance = $\sqrt{{\left(1\right)}^2+{\left(-7\right)}^2}$

Solving further

Distance = $\sqrt{1+49}$

The distance is

  $\sqrt{50}=7.07$

Now, take J (-2, 4) and L $\left(-\frac{5}{2},\frac{1}{2}\right)$

Substitute values in the formula

Distance = $\sqrt{{\left(-\frac{5}{2}+2\right)}^2+{\left(\frac{1}{2}-4\right)}^2}$

Solving it further using lower common factor method

Distance = $\sqrt{{\left(\frac{-5+4}{2}\right)}^2+{\left(\frac{1-8}{2}\right)}^2}$

Solving it further

Distance = $\sqrt{{\left(\frac{-1}{2}\right)}^2+{\left(\frac{-7}{2}\right)}^2}$

Solving further

  $\sqrt{\frac{1}{4}+\frac{49}{4}}=\sqrt{\frac{50}{4}}$

The distance is

  $3.53$

Now, take K (-3, -3) and L $\left(-\frac{5}{2},\frac{1}{2}\right)$

Substitute the values in the formula

Distance = $\sqrt{{\left(-\frac{5}{2}+3\right)}^2+{\left(\frac{1}{2}+3\right)}^2}$

Solving using lowest common factor inside the bracket

  $\sqrt{{\left(\frac{-5+6}{2}\right)}^2+{\left(\frac{1+6}{2}\right)}^2}$

Solving further

  $\sqrt{\frac{1}{4}+\frac{49}{4}}=\sqrt{\frac{50}{4}}$

The distance is

  $3.53$