Question

In a triangle the two vertices are $(0,0),(3,4).$ If orthocentre of the triangle is $(1,4)$ then third vertex of the triangle is:

• A. $(1,-2)$
• B. $(\frac{4}{19},0)$
• C. $(0,\frac{19}{4})$
• D. $(-\frac{4}{19},\frac{19}{4})$

Answer 1

The correct option is C $(0,\frac{19}{4})$

$A(0,0),B(3,4),0(1,4)$ and $C(x,y)$
As O is the orthocenter of $△ABC$
$OC\perp AB\Rightarrow \frac{4-y}{1-x}=\frac{4-0}{3-0}\Rightarrow 3(4-y)=4(1-x)\Rightarrow 4x-3y=-8$ ...(1)
And $OA\perp BC\Rightarrow \frac{4-0}{1-0}=\frac{4-y}{3-x}\Rightarrow 4x-y=8$ ...(2)
Solving (1) and (2), we get
$x=0,y=\frac{19}{4}$
$\therefore (0,\frac{19}{4})=(x,y)$