a) The slope of line CA (or if you'd like, a line that exactly overlays CA) is () = (8-3)/(0-(-1)) = 5/1 = 5  and for a line overlaying CB, the slope  = (8-7)/(0-5) = -1/5.




When two lines have slopes such that  they are perpendicular.

Since this is true for CA and CB, those legs of the triangle are perpendicular and the angle ACB is a right angle.






An alternate method is to treat CA and CB as vectors (take the "tip" and subtract the "tail"):

CA = <-1-0, 3-8> = <-1,-5>

CB = <5-0, 7-8> = <5,-1>



and then take the dot product: -1*5 + -5*-1 = -5 + 5 = 0.  This again shows CA perpendicular to CB because (only) when two nonzero vectors are orthogonal (at 90 degrees to each other), will they have a dot product of zero.   




One problem per post... but here is a head-start on (b)




b) We found slope AC in part (a).  It was 5.   So the equation of a line through B with slope 5 is:  y-7 = 5(x-5)



This line crosses the x-axis when y=0. You therefore need to set y=0 and solve for x. 

 

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