Question 1167021
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a) The slope of line CA (or if you'd like, a line that exactly overlays CA) is ({{{m[1]}}}) = (8-3)/(0-(-1)) = 5/1 = 5  and for a line overlaying CB, the slope {{{ m[2] }}} = (8-7)/(0-5) = -1/5.
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When two lines have slopes such that {{{m[1] = -1/m[2] }}} 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.

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