A triangle has corners A, B, and C located at #(2 ,2 )#, #(3 ,4 )#, and #(6 ,8 )#, 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  ,2   )#, #(3  ,4    )#, and #(6 ,8   )#, respectively. What are the endpoints and length of the altitude going through corner C?

Ella Armstrong · Accepted Answer

#"The endpoints of altitude through corner C are:"C(6,8) " and "D(5.2,8.4)#

#"The length of altitude is : "0.89#

#"The slope of line through A(2,2) and B(3,4) is :" #

#m=(4-2)/(3-1)=2/1=2#

#"now,let us find the slope of line passing C(6,8) and D(x,y)"#

#"if two line is perpendicular , product of their slopes is equal to -1"#

#m*n=-1#
#2*n=-1#
#n=-1/2#

#"now ,let us write equations of lines "#

#"equation for line passing A(2,2) and D(x,y)"#
#y-2=2(x-2)#
#y=2x-4+2#
#y=2x-2" (1)"#

#"equation for line passing C(6,8) and D(x,y)"#
#y-8=-1/2(x-6)#
#y=-1/2 (x-6)+8#

#y=-1/2 x+3+8#

#y=-1/2 x+11" (2)"#

#"let us write the equations (1) and (2) as equal "#

#2x-2=-1/2 x+11#

#2x+1/2 x=11+2#

#5/2 x=13" , "5x=13*2" ," x=26/5" , "x=5.2#

#"use equation (1) or (2)"#

#y=2x-2#

#"insert x=5.2"#

#y=2*5.2-2#
#y=10.4-2#
#y=8.4#

#D=(5.2,8.4)#

#"the end points are C(6,8) and D(5.2,8.4)"#

#"The length of the altitude can be calculated using :"#

#h_C=sqrt((5.2-6)2+(8.4-8)^2)#

#h_C=sqrt((-0.8)+(0.4)^2)#

#h_C=sqrt(0.64+0.16)#

#h_C=sqrt(0.80)#

#h_C=0.89#

Mason Adams · Answer

undefined To find the endpoints and length of the altitude going through corner C of the triangle with vertices A(2, 2), B(3, 4), and C(6, 8), we first need to determine the slope of the line segment AB and then find the perpendicular slope to this segment, which will be the slope of the altitude.

The slope of AB is given by:
$$ m_{AB} = \frac{y_B - y_A}{x_B - x_A} $$

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

The perpendicular slope to AB, which is the slope of the altitude, is the negative reciprocal of the slope of AB:
$$ m_{\text{altitude}} = -\frac{1}{m_{AB}} = -\frac{1}{2} $$

Now that we have the slope of the altitude, we can use point-slope form to find the equation of the line passing through C(6, 8) with slope -1/2:
$$ y - y_C = m_{\text{altitude}}(x - x_C) $$
$$ y - 8 = -\frac{1}{2}(x - 6) $$
$$ y - 8 = -\frac{1}{2}x + 3 $$
$$ y = -\frac{1}{2}x + 11 $$

Next, we find the intersection point of this line with the line segment AB. Substitute the x-value of this point into the equation of the altitude to find the corresponding y-value:
$$ -\frac{1}{2}x + 11 = 2x - 2 $$
$$ \frac{5}{2}x = 13 $$
$$ x = \frac{26}{5} $$

Substitute $ x = \frac{26}{5} $ into the equation of the altitude to find y:
$$ y = -\frac{1}{2}(\frac{26}{5}) + 11 = \frac{49}{5} $$

Therefore, the endpoints of the altitude going through corner C are $ (\frac{26}{5}, \frac{49}{5}) $ and $ (6, 8) $. To find the length of the altitude, use the distance formula between these endpoints:
$$ \text{Length} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$
$$ \text{Length} = \sqrt{(\frac{26}{5} - 6)^2 + (\frac{49}{5} - 8)^2} $$
$$ \text{Length} = \sqrt{(\frac{16}{5})^2 + (\frac{9}{5})^2} $$
$$ \text{Length} = \sqrt{\frac{256}{25} + \frac{81}{25}} $$
$$ \text{Length} = \sqrt{\frac{337}{25}} $$
$$ \text{Length} = \frac{\sqrt{337}}{5} $$

So, the endpoints of the altitude are $ (\frac{26}{5}, \frac{49}{5}) $ and $ (6, 8) $, and the length of the altitude is $ \frac{\sqrt{337}}{5} $.