Question

Write the coordinates of the orthocentre of the triangle by points (8,0),(4,6) and (0,0).

Answer 1

$\text{Mark the points in the graph as follows:}$

$\text{Let us find the distance AB and BC}SinceAB=\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}andBC=\sqrt{(8-4{)}^2+(0-6{)}^2}=\sqrt{4^2+(-6{)}^2}=\sqrt{52}=2\sqrt{13}\Delta ABC\text{is an isosceles triangle with equal sides AB and BC.}\text{The orthocentre of an isosceles triangle lies on the altitude from the vertex B to the base AC and it bisects AC.}\text{Let D be the foot of the perpendicular from B to AC.}\text{Thus,the coordinates of D are (4,0) since AC =8 units}\text{Thus, equation of BD is x=4~~~~~~...(1)}\text{Now let us find the equation of the side BC BC is the line joining the points B(4,6) and C(8,0)}ThusequationofBCis\frac{y-y_2}{y_2-y_1}=\frac{x-x_2}{x_2-x_1}\Rightarrow equationofBCis\frac{y-0}{0-6}=\frac{x-8}{8-4}\Rightarrow equationofBCis\frac{y}{-6}=\frac{x-8}{4}\Rightarrow equationofBCis4y=-6x+48\Rightarrow equationofBCisy=-\frac{3}{2}x+48\text{Now the slope of the perpendicular from the vertex A to the base BC is}-\frac{1}{(-\frac{3}{2})}=\frac{2}{3}\text{Thus,the equation of the line AE,having slope}\frac{2}{3}\text{and passing through the origin is y}=\frac{2}{3}x....\left(2\right)\text{Now let us find the intersection of the lines BD and AE}\text{Thus from equation (1) and (2),we have}y=\frac{2}{3}\times 4=\frac{8}{3}.\text{Thus the intersection points O, the irthocentre is having coordinates}O(4,\frac{8}{3})$