   Chapter 10.5, Problem 45E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In Exercises 45 and 46, complete an analytic proof for each theorem.The altitudes of a triangle are concurrent.

To determine

To prove:

The altitudes of triangle are concurrent.

Explanation

The triangle ABC with vertices A0, 0, Ba, 0, and Cb, c.is shown in the above figure.

Let AJ-, BK-, CH- be the altitudes of the triangle ABC.

If the point of intersection of the altitudes is at the same point, then they are said to be concurrent.

By theorem,

If two lines are perpendicular, then the product of their slopes is -1 or one slope is negative reciprocal of the other slope.

(i.e.) If l1l2, then m1.m2=-1 or m2=-1m1

Since AB- is the horizontal line, slope is zero.

(i.e.) mAB-=0

Obviously the altitude CH- is vertical and it has the equation,

x=b

The slope of the line BC- with vertices Ba, 0, and Cb, c is as follows.

Using the slope formula and choosing x1=a, x2=b, y1=0, and y2=c.

mBC-=c-0b-a

mBC-=cb-a

Thus the slope of the altitude AJ- which is perpendicular to the line BC- is

mAJ-=-b-ac=a-bc

The equation of the line AJ- that contains the point 0, 0 with slope a-bc is as follows:

Initially substitute 0, 0 in the equation y=mx+B,

0=a-bc0+B

B=0

Substitute B=0 in the equation y=mx+B,

y=a-bcx

The intersection of these altitudes is found by solving the two altitude equation.

x=b and y=a-bcx

On solving,

y=a-bcb

y=ab-b2c

The coordinates of orthocenter x, y is b, ab-b2c

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