Question

The orthocentre of the triangle formed by \[\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \right)\] is
A) $\left( {4,\dfrac{8}{3}} \right)$
B) $(3,4)$
C) $(4,3)$
D) $( - 3,4)$

Answer 1

Hint: For solving this particular question, we must know that slope of a line can also be found if two points on the line are given . let the two points on the line be $({x_1},{y_1}),({x_2},{y_2})$ respectively.
Then the slope is given by , $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ .
Slope is also defined as the ratio of change in $y$ over the change in $x$between any two points.

Complete step-by-step solution:
It is given that \[\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \right)\] are the vertices of a triangle.
Let ABC be a triangle having the vertices \[\left( {0,0} \right),\left( {8,0} \right),\left( {4,6} \right)\] ,
Let ‘A’ be a vertex \[\left( {0,0} \right)\] ,
‘B’ be a vertex \[\left( {8,0} \right)\] ,
And ‘C’ be a vertex \[\left( {4,6} \right)\] .
Slope of a line can also be found if two points on the line are given . let the two points on the line be $({x_1},{y_1}),({x_2},{y_2})$ respectively.
Then slope is given by , $m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$ .
Slope is also defined as the ratio of change in $y$ over the change in $x$between any two points.
Therefore, slope of $BC = \dfrac{{6 - 0}}{{4 - 8}} = - \dfrac{2}{3}$ .
Now, we have to find equation of the line through ‘A’ which is perpendicular to $BC$ is ,
$
   \Rightarrow y - 0 = \left( {\dfrac{2}{3}} \right)(x - 0) \\
   \Rightarrow 2x - 3y = 0......................(1) \\
 $
Now, slope of $CA = \dfrac{{0 - 6}}{{0 - 4}} = \dfrac{3}{2}$ .
Now, we have to find equation of the line through ‘B’ which is perpendicular to $CA$ is ,
$
   \Rightarrow y - 0 = \left( { - \dfrac{3}{2}} \right)(x - 8) \\
   \Rightarrow 2x + 3y = 16.....................(2) \\
 $
Now, we have to solve $(1),(2)$ , after solving we will get the orthocentre as $\left( {4,\dfrac{8}{3}} \right)$ .

Therefore, option ‘A’ is the correct option.

Note: This type of linear equations sometimes called slope-intercept form because we can easily find the slope and the intercept of the corresponding lines. This also allows us to graph it. We can quickly tell the slope i.e., $m$ the y-intercepts i.e., $(y,0)$ and the x-intercept i.e., $(0,y)$ .we can graph the corresponding line .