A triangle has corners A, B, and C located at #(4 ,8 )#, #(7 ,4 )#, and #(5 ,3 )#, respectively. What are the endpoints and length of the altitude going through corner C?

Let CD be the altitude drawn from C perpendicular to side AB touches at D of side AB. Slope of line AB is #(4-8)/(7-4) = -4/3 :.#slope of line CD is # -1/(-4/3) =3/4# Equation of line CD is #y-3=3/4(x-5) or 4y-12=3x-15 or 4y-3x= -3; (1)# Equation of line AB is #y-8 = -4/3(x-4) or 3y-24 =-4x +16 or 3y+4x = 40; (2) # Now solving equation (1) & equation (2) we find the co-ordinate of poin D. #12y-9x = -9 # (1)x3 (3) #12y+16x = 160# (2)x4 (4) subtracting (3) from (4) we get #25x=169 or x =169/25=6.76 :. y =(40-(4*6.76))/3 =4.32# End points of the altitude are #(5,3) & (6.76,4.32)# Length of altitude #= sqrt((5-6.76)^2+(3-4.32)^2) =2.2#[Ans]

To find the endpoints and length of the altitude from corner C to the line formed by points A and B, follow these steps: ### Slope of AB First, find the slope of line AB. The slope \(m\) is given by \[m = \frac{y_2 - y_1}{x_2 - x_1}\] Substituting the coordinates of A(4, 8) and B(7, 4), \[m_{AB} = \frac{4 - 8}{7 - 4} = \frac{-4}{3}\] ### Slope of the Altitude from C The altitude from point C to line AB is perpendicular to AB, so its slope \(m_{\text{altitude}}\) is the negative reciprocal of \(m_{AB}\), \[m_{\text{altitude}} = -\frac{1}{m_{AB}} = \frac{3}{4}\] ### Equation of the Altitude With the slope of the altitude and the point C(5, 3), use the point-slope form of the equation of a line: \[y - y_1 = m(x - x_1)\] \[y - 3 = \frac{3}{4}(x - 5)\] \[y = \frac{3}{4}x + \frac{3}{4}\cdot(-5) + 3\] \[y = \frac{3}{4}x - \frac{15}{4} + \frac{12}{4}\] \[y = \frac{3}{4}x - \frac{3}{4}\] ### Intersection with Line AB Line AB can be expressed using the two points A and B. However, to simplify calculations and directly find the intersection, we focus on the altitude's intersection with AB. Since we have the slope-intercept form of the altitude, we find the equation of line AB to intersect them. To find the equation of AB, use the slope \(-\frac{4}{3}\) and point A(4, 8): \[y - 8 = -\frac{4}{3}(x - 4)\] \[y = -\frac{4}{3}x + \frac{16}{3} + 8\] \[y = -\frac{4}{3}x + \frac{16}{3} + \frac{24}{3}\] \[y = -\frac{4}{3}x + \frac{40}{3}\] ### Solving for Intersection Now, solve for the intersection of \(y = \frac{3}{4}x - \frac{3}{4}\) and \(y = -\frac{4}{3}x + \frac{40}{3}\) by setting them equal: \[\frac{3}{4}x - \frac{3}{4} = -\frac{4}{3}x + \frac{40}{3}\] Multiplying every term by 12 to clear the denominators: \[9x - 9 = -16x + 160\] \[25x = 169\] \[x = \frac{169}{25}\] Plugging \(x\) back into the equation of the altitude to find \(y\): \[y = \frac{3}{4}\left(\frac{169}{25}\right) - \frac{3}{4} = \frac{507}{100} - \frac{75}{100} = \frac{432}{100} = \frac{108}{25}\] ### Endpoints and Length of the Altitude The endpoints of the altitude are C(5, 3) and the point of intersection \(\left(\frac{169}{25}, \frac{108}{25}\right)\). To find the length \(d\) of the altitude, \[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] \[d = \sqrt{\left(\frac{169}{25} - 5\right)^2 + \left(\frac{108}{25} - 3\right)^2}\] \[d = \sqrt{\left(\frac{169}{25} - \frac{125}{25}\right)^2 + \left(\frac{108}{25} - \frac{75}{25}\right)^2}\] \[d = \sqrt{\left(\frac{44}{25}\right)^2 + \left(\frac{33}{25}\right)^2}\] \[d = \sqrt{\frac{1936}{625} + \frac{1089}{625}}\] \[d = \sqrt{\frac{3025}{625}}\] \[d = \frac{55}{25} = \frac{11}{5}\] The endpoints of the altitude are \(C(5, 3)\) and approximately \(\left(6.76, 4.32\right)\), and its length is \(\frac{11}{5}\) or 2.2 units.