This exercise illustrates the fact that the altitudes of a triangle are concurrent. Agai...

This exercise illustrates the fact that the altitudes of a triangle are concurrent. Again, we’ll be using with vertices A(-4, 0), B(2, 0), and C(0, 6). Note that one of the altitudes of this triangle is just the portion of the y-axis extending from y = 0 to y = 6; thus, you won’t need to graph this altitude; it will already be in the picture.

(a) Using paper and pencil, find the equations for the three altitudes. (Actually, you are finding equations for the lines that coincide with the altitude segments.)

(b) Use a graphing utility to draw along with the three altitude lines that you determined in part (a). Note that the altitudes appear to intersect in a single point. Use the graphing utility to estimate the coordinates of this point.

(c) Using simultaneous equations (from intermediate algebra), find the exact coordinates of the orthocenter. Are your estimates in part (b) close to these values?