Question

The orthocentre of the triangle with vertices $ \left( {0,0} \right),\,\left( {3,4} \right)\,\,and\,\left( {4,0} \right) $ is:
 $
  A.\,\,\left( {3,\dfrac{5}{4}} \right) \\
  B.\,\,\left( {3,12} \right) \\
  C.\,\,\left( {3,\dfrac{3}{4}} \right) \\
  D.\,\,\left( {3,9} \right) \\
  $

Answer 1

Hint: To find coordinate of orthocentre of triangle. We first find the equation of altitudes using the slope of the opposite line on which altitudes are drawn and the point through which it passes and then solve these equations together to find their point of intersection or orthocentre of the triangle with given vertices.
Formulas used: Equation of a line through two points and having slope is given as:
 $ y - {y_1} = m\left( {x - {x_1}} \right) $ , slope of a line passing through two points given as: $ \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} $

Complete step-by-step answer:
Given vertices of triangle are $ \left( {0,0} \right),\,\left( {3,4} \right)\,\,and\,\left( {4,0} \right) $
Let $ A\left( {0,0} \right),B\left( {3,4} \right)\,\,and\,\,C\left( {4,0} \right) $ are three vertices of triangle ABC.
Let AD, BE and CF are altitudes from each vertex of triangle ABC and O is the intersection point of all altitudes or we can say orthocentre of triangle ABC.

Slope of line AB passing through two points $ A(0,0)\,\,and\,\,B\left( {3,4} \right) $ is given as:
 $
  \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} = \dfrac{{4 - 0}}{{3 - 0}} \\
   = \dfrac{4}{3} \;
  $
Since, line CF is perpendicular to line AB.
Therefore, slope of line CF will be negative of reciprocal of line AB.
Hence, slope of CF will be equal to $ - \dfrac{3}{4} $
As, altitude passing through point $ C\left( {4,0} \right) $ and having slope $ - \dfrac{3}{4} $ . Therefore, its equation will be given as:
 $ y - {y_1} = m\left( {x - {x_1}} \right) $
Substituting value in above equation. We have,
 $
  y - 0 = - \dfrac{3}{4}\left( {x - 4} \right) \\
   \Rightarrow y = - \dfrac{3}{4}\left( {x - 4} \right) \\
   \Rightarrow 4y = - 3x + 12 \\
   \Rightarrow 3x + 4y = 12 \;
  $
Therefore, equation of altitude CF is $ 3x + 4y = 12 $
Now, we will find the equation of altitude AD.
Slope of line BC given as:
 $ \dfrac{{0 - 4}}{{4 - 3}} = \dfrac{{ - 4}}{1} $
As, AD is perpendicular to line BC. Therefore, the slope of AD will be equal to $ \dfrac{1}{4} $ .
Equation of line AD will be given as:
 $
  y - 0 = \dfrac{1}{4}\left( {x - 0} \right) \\
   \Rightarrow y = \dfrac{1}{4}x \;
  $
Since, we know that orthocentre is a point of intersection of altitudes. Therefore, solving above two equations of altitude we will get their point of intersection or coordinate of orthocentre.
Above formed equation of altitudes are $ 3x + 4y = 12\,\,\,\,and\,\,\,y = \dfrac{1}{4}x $
Substituting value of y from second equation in first equation.
 $
  3x + 4\left( {\dfrac{1}{4}x} \right) = 12 \\
   \Rightarrow 3x + x = 12 \\
   \Rightarrow 4x = 12 \\
   \Rightarrow x = \dfrac{{12}}{4} \\
  x = 3 \;
  $
Substituting value of x calculated above to find value of y.
 $
  y = \dfrac{1}{4}(3) \\
  or \\
  y = \dfrac{3}{4} \;
  $
Therefore, from above we see that point of intersection of altitude is $ \left( {3,\dfrac{3}{4}} \right) $
Hence, the coordinate of the orthocentre of triangle ABC is $ \left( {3,\dfrac{3}{4}} \right) $ .
So, the correct answer is “Option C”.

Note: Orthocentre is a point where all altitudes of the triangle meet. Therefore, we can find coordinates of the orthocenter by solving equations of altitudes. As, triangle has three altitudes. Therefore, there will be three equations but it requires only two to find the coordinate of intersection. So, we can use either of two equations or we form two equations to find coordinates of the orthocentre of a triangle with given vertices.