A triangle has corners A, B, and C located at #(2 ,3 )#, #(8 ,8 )#, and #(4 , 2 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Question

A triangle has corners A, B, and C located at #(2   ,3   )#, #(8  ,8    )#, and #(4 ,  2   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Violet Aldrich · Accepted Answer

The end points are #=(164/61,654/183)# and the length is #=2.05u#

The triangle's corners are

#A=(2,3)#

#B=(8,8)#

#C=(4,2)#

The slope of the line #AB# is #m=(8-3)/(8-2)=5/6#

The equation of line #AB# is

#y-3=5/6(x-2)#

#y-3=5/6x-5/3#

#y=5/6x+4/3#...........................#(1)#

#mm'=-1#

The slope of the line perpendicular to #AB# is #m'=-6/5#

The equation of the altitude through #C# is

#y-2=-6/5(x-4)#

#y-2=-6/5x+24/5#

#y=-6/5x+34/5#................................#(2)#

Solving for #x# and #y# in equations #(1)# and #(2)#, we get

#5/6x+4/3=-6/5x+34/5#

#5/6x+6/5x=34/5-4/3#

#61/30x=82/15#

#x=164/61#

#y=5/6*164/61+4/3=410/183+4/3=654/183#

The end points of the altitude is #=(164/61,654/183)#

The altitude's duration is

#=sqrt((4-164/61)^2+(2-654/183)^2)#

#=sqrt((1.311)^2+(-1.574)^2)#

#=sqrt(4.197)#

#=2.05#

Theodore Abad · Answer

undefined To find the altitude going through corner C in the triangle with vertices A(2, 3), B(8, 8), and C(4, 2), we first need to find the equation of the line containing side AB, as the altitude through C will be perpendicular to AB.

The slope of the line AB is given by:

$$ m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{8 - 3}{8 - 2} = \frac{5}{6} $$

The perpendicular slope to $ m_{AB} $ will be the negative reciprocal of $ m_{AB} $, so the slope of the altitude through C, denoted as $ m_{altitude} $, is $ -\frac{6}{5} $.

Using the point-slope form of a line equation $ y - y_1 = m(x - x_1) $ with the point $ (x_1, y_1) = (4, 2) $, we can find the equation of the altitude:

$$ y - 2 = -\frac{6}{5}(x - 4) $$
$$ y = -\frac{6}{5}x + \frac{24}{5} + 2 $$
$$ y = -\frac{6}{5}x + \frac{34}{5} $$

Now, to find the endpoints of the altitude, we need to find the intersection points of this line with the lines containing sides AC and BC.

For side AC, we can find its equation and solve the system of equations. The slope of AC is:

$$ m_{AC} = \frac{y_C - y_A}{x_C - x_A} = \frac{2 - 3}{4 - 2} = -\frac{1}{2} $$

Using point-slope form with point $ (x_1, y_1) = (4, 2) $:

$$ y - 2 = -\frac{1}{2}(x - 4) $$
$$ y = -\frac{1}{2}x + 4 $$

To find the intersection point, we solve:

$$ -\frac{6}{5}x + \frac{34}{5} = -\frac{1}{2}x + 4 $$

Solving this equation yields $ x = \frac{76}{31} $. Substituting this value into either equation (let's use the equation of the altitude line), we find $ y = \frac{6}{5}(\frac{76}{31}) + \frac{34}{5} = \frac{294}{31} $.

So, the endpoints of the altitude are approximately $ \left(\frac{76}{31}, \frac{294}{31}\right) $ and $ (4, 2) $.

Finally, to find the length of the altitude, we use the distance formula:

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

Substituting the coordinates of the two endpoints, we get:

$$ d = \sqrt{\left(\frac{76}{31} - 4\right)^2 + \left(\frac{294}{31} - 2\right)^2} $$

After calculation, the length of the altitude is approximately $ \frac{12\sqrt{221}}{31} $ units.