Question

If the point \(M\) is the foot of the perpendicular drawn from the point \(P(0,9)\) to the straight line \[ 2x-5y+16=0, \] and \(Q=(12,8)\), then the orthocentre of \(\triangle MPQ\) is

• A. \((2,4)\)
• B. \(\left(6,\dfrac{17}{2}\right)\)
• C. \((7,6)\)
• D. \(\left(\dfrac{14}{3},7\right)\)

Hint:

In any right-angled triangle, the orthocentre coincides with the vertex at which the right angle is formed.

Answer 1

The Correct Option is A

Step 1: Find the foot of the perpendicular \(M\). The given line is \[ 2x-5y+16=0. \] Using the foot of perpendicular formula from \(P(0,9)\), \[ M= \left( x_1-\frac{a(ax_1+by_1+c)}{a^2+b^2}, \; y_1-\frac{b(ax_1+by_1+c)}{a^2+b^2} \right). \] Here, \[ a=2,\; b=-5,\; c=16, \] and \[ ax_1+by_1+c = 2(0)-5(9)+16 = -29. \] Hence, \[ M= \left( 0-\frac{2(-29)}{29}, \; 9-\frac{-5(-29)}{29} \right) = (2,4). \]

Step 2: Use the property of a right triangle. Since \(PM\) is perpendicular to the line containing \(MQ\), \[ PM\perp MQ. \] Thus, \[ \angle PMQ=90^\circ. \] Therefore, \(\triangle MPQ\) is a right-angled triangle at \(M\).

Step 3: Identify the orthocentre. In a right-angled triangle, the orthocentre is the vertex containing the right angle. Hence, \[ \boxed{\text{Orthocentre}=M=(2,4).} \] Therefore, the correct option is \[ \boxed{(A)\ (2,4).} \]