A triangle has corners A, B, and C located at #(6 ,1 )#, #(2 ,3 )#, and #(3 , 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 #(6   ,1   )#, #(2  ,3    )#, and #(3 ,  2   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Allison Allen · Accepted Answer

#"End points "(3,2) & (16/5,12/5)#

#"Altitude"=1/sqrt5# Given the coordinates of three corners of #DeltaABC# #A->(6,1)# #B->(2,3)# #C->(3,2)# The perpendicular dropped from C on AB at D is denoted by CD. Our task is to determine the coordinate of D and the length of CD. Let (h,k) be the coordinate of D. #"Now slope of AB"=(1-3)/(6-2)=-1/2# Thus, CD's slope perpendicular to AB is 2. However, the slope CD with respect to its endpoints (h,k) and (3,2) is #(k-2)/(h-3)# So #(k-2)/(h-3)=2# #=>2h-6=k-2# #=>2h-k=4.....(1)# Equation AB is now #y-3=-1/2(x-2)# #2y-6=-x+2# #x+2y=8# The solution to this equation is (h,k). So#" "h+2k=8....(2)# Now, by multiplying (1) by 2, adding (2), and so forth, we obtain #4h-2k+h+2k=2*4+8# #5h=16=>h=16/5# Entering h's value in (1) #2*16/5-k=4# #k=32/5-4=12/5# Thus, D's coordinate is #D->(16/5,12/5)# How long is the altitude CD? #=sqrt((16/5-3)^2+(12/5-2)^2)# #=sqrt((1/5)^2+(2/5)^2)# #=sqrt(1/25+4/25)# #=sqrt(5/25)# #=1/sqrt5#

Noah Atkinson · Answer

undefined To find the endpoints and length of the altitude going through corner C, first, determine the slope of side AB, then use the negative reciprocal of that slope to find the slope of the altitude. Next, use the point-slope form to find the equation of the altitude passing through point C. After finding the equation, solve it simultaneously with the equation of side AB to find the intersection point, which is the endpoint of the altitude. Finally, calculate the distance between point C and the endpoint to find the length of the altitude.

Given the coordinates of points A (6, 1) and B (2, 3), the slope of side AB is $ m_{AB} = \frac{y_B - y_A}{x_B - x_A} $.

$$ m_{AB} = \frac{3 - 1}{2 - 6} = \frac{2}{-4} = -\frac{1}{2} $$

The negative reciprocal of $ m_{AB} $ is the slope of the altitude, $ m_{\text{altitude}} $.

$$ m_{\text{altitude}} = -\frac{1}{m_{AB}} = -\frac{1}{-\frac{1}{2}} = 2 $$

Using the point-slope form of a line with corner C (3, 2) and the slope $ m_{\text{altitude}} $, the equation of the altitude is:

$$ y - y_C = m_{\text{altitude}}(x - x_C) $$

$$ y - 2 = 2(x - 3) $$

$$ y - 2 = 2x - 6 $$

$$ y = 2x - 4 $$

Now, to find the intersection point of this line with side AB, you solve the system of equations formed by the equation of side AB and the equation of the altitude.

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

Solve the system:

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

$$ \frac{5}{2}x = 8 $$

$$ x = \frac{16}{5} $$

Substitute $ x = \frac{16}{5} $ into the equation of side AB:

$$ y = -\frac{1}{2} \times \frac{16}{5} + 4 $$

$$ y = -\frac{8}{5} + 4 = \frac{12}{5} $$

So, the endpoint of the altitude going through corner C is $ \left(\frac{16}{5}, \frac{12}{5}\right) $.

Now, calculate the distance between point C and this endpoint using the distance formula:

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

$$ d = \sqrt{\left(\frac{16}{5} - 3\right)^2 + \left(\frac{12}{5} - 2\right)^2} $$

$$ d = \sqrt{\left(\frac{16}{5} - \frac{15}{5}\right)^2 + \left(\frac{12}{5} - \frac{10}{5}\right)^2} $$

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

$$ d = \sqrt{\frac{1}{25} + \frac{4}{25}} $$

$$ d = \sqrt{\frac{5}{25}} $$

$$ d = \frac{\sqrt{5}}{5} $$

So, the length of the altitude going through corner C is $ \frac{\sqrt{5}}{5} $ units.