document.write( "Question 1167021: A triangle is formed by the points A( - 1, 3), B(5, 7) and C(0, 8).
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document.write( "a) Show that the angle ACB is a right angle.
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document.write( "b) Find the coordinate of the point where the line through B parallel to AC cuts the x-axis. \n" );
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Algebra.Com's Answer #791583 by math_helper(2461)  You can put this solution on YOUR website! \r \n" );
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document.write( "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. \n" );
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document.write( "When two lines have slopes such that they are perpendicular. \n" );
document.write( "Since this is true for CA and CB, those legs of the triangle are perpendicular and the angle ACB is a right angle.\r \n" );
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document.write( "An alternate method is to treat CA and CB as vectors (take the \"tip\" and subtract the \"tail\"): \n" );
document.write( "CA = <-1-0, 3-8> = <-1,-5> \n" );
document.write( "CB = <5-0, 7-8> = <5,-1>\r \n" );
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document.write( "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. \n" );
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document.write( "One problem per post... but here is a head-start on (b) \n" );
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document.write( "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)\r \n" );
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document.write( "This line crosses the x-axis when y=0. You therefore need to set y=0 and solve for x. \n" );
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