Find the coordinates of the orthocenter of ΔYAB that has vertices at Y(3, –2), A(3, 5), and B(9, 1). (JUSTIFY)

Accepted Solution

Answer:So (5,1) is the orthocenter. Step-by-step explanation:So we have to find the slopes of all three lines in burgundy (the line segments of the triangle). We also need to find the equations for the altitudes with respect from all sides of the triangle (we are looking for perpendicular lines).The vertical line there is just going to be x=a number so that line is x=3 because all the points on that line are of the form (3,y).  x=3 says we don't care what y is but x will always be 3.  So the line for AY is x=3.So the altitude of the triangle with respect to that side (that line segment) would be a line that is perpendicular to is which would be a horizontal line y=1.  I got y=1 because it goes through vertex B(9,1) and y=1 is perpendicular to x=3.  So we now need to find the equations of the other 2 lines.One line has points A(3,5) and B(9,1).To find the slope, you may use [tex]\frac{y_2-y_1}{x_2-x_1}[/tex].Or you could just line up the points vertically and subtract then put 2nd difference over 1st difference.Like this: (9 , 1)-(3  ,5)---------- 6     -4So the slope is -4/6=-2/3.So a line that is perpendicular will have opposite reciprocal slope.  That means we are looking for a line with 3/2 as the slope.  We want this line from segment AB going to opposite point Y so this line contains point (3,-2).Point slope-form is[tex]y-y_1=m(x-x_1)[/tex] where [tex](x_1,y_1)[/tex] is a point on the line and [tex]m[/tex] is the slope.So the line is:[tex]y-(-2)=\frac{3}{2}(x-3)[/tex][tex]y+2=\fac{3}{2}x-\frac{9}{2}[/tex]Subtract 2 on both sides:[tex]y=\frac{3}{2}x-\frac{9}{2}-2[/tex]Simplify:[tex]y=\frac{3}{2}x-\frac{13}{2}[/tex].Let's find the the third line but two lines is plenty, really. The othorcenter is where that perpendicular lines will intersect. Now time for the third line.BY has points (9,1) and (3,-2).The slope can be found by lining up the points vertically and subtracting, then put 2nd difference over 1st difference: (9  ,1)-(3,-2)---------6     3So the slope is 3/6=1/2.A perpendicular line will have opposite reciprocal slope. So the perpendicular line will have a slope of -2.We want this line segment to go through A(3,5).We are going to use point-slope form:[tex]y-5=-2(x-3)[/tex]Add 5 on both sides:[tex]y=-2(x-3)+5[/tex]Distribute:[tex]y=-2x+6+5[/tex]Combine like terms:[tex]y=-2x+11[/tex]So the equation of the 3rd altitude line is y=-2x+11.So the equations we want to find the intersection to is:y=(3/2)x-(13/2)y=1y=-2x+11I like the bottom two equations so I'm going to start there and then use my third line to check some of my work.y=1y=-2x+11Replacing 2nd y with 1 since y=1:1=-2x+11Subtract 11 on both sides:1-11=-2xSimplify:-10=-2xDivide both sides by -2:5=xThe point of intersection between y=1 and y=-2x+11 is (5,1).Let's see if (5,1) is on that third line.y=(3/2)x-(13/2)1=(3/2)(5)-(13/2)1=(15/2)-(13/2)1=(2/2)1=1So (5,1) is the intersection of all three lines.So (5,1) is the orthocenter.