Chapter P.2, Problem 55E

Solution

Prove that the midpoint of the hypotenuse of the right triangle with vertices (0, 0), (5, 0), and (0, 7) is equidistant from the three vertices.

AD=CD=BD= $\sqrt{18.5}$

The Mid-point of right triangle is equidistant from all the three vertices.

Given information:

Vertices of right triangle are (0, 0), (5, 0), and (0, 7)

Concept used:

• First, find the value for x on the x-axis.
• Next, find the y-value on the y-axis.
• The point should be plotted at the intersection of x and y.
• Finally, plot the point on your graph at the appropriate spot.
• Then join all the points to make a closed figure.

Calculation:

The given points are A(0, 0), B(5, 0), and C(0, 7)

Plotting the points in the graph -

  

The mid-point is D(2.5, 3.5).

Using distance formula −

A(0, 0), D(2.5, 3.5)

  $\begin{array}{c}AD=\sqrt{(x_1-x_2)+(y_1-y_2)}\\ =\sqrt{{(0-2.5)}^2+{(0-3.5)}^2}\\ =\sqrt{6.25+12.25}\\ =\sqrt{18.5}\end{array}$

C(0,7), D(2.5, 3.5)

  $\begin{array}{c}CD=\sqrt{(x_1-x_2)+(y_1-y_2)}\\ =\sqrt{{(0-2.5)}^2+{(7-3.5)}^2}\\ =\sqrt{{(-2.5)}^2+{(3.5)}^2}\\ =\sqrt{6.25+12.25}\\ =\sqrt{18.5}\end{array}$

B(5, 0), D(2.5, 3.5)

  $\begin{array}{c}BD=\sqrt{(x_1-x_2)+(y_1-y_2)}\\ =\sqrt{{(5-2.5)}^2+{(0-3.5)}^2}\\ =\sqrt{6.25+12.25}\\ =\sqrt{18.5}\end{array}$

AD=CD=BD= $\sqrt{18.5}$

As D is the mid-point of hypotenuse CB and the distance from three vertices to mid-point is same.

So mid-point of right triangle is equidistant from all the three vertices.