Question

Two vertices of a triangle are $(5,-1)$ and $(-2,3)$. If the origin is the orthocentre of this triangle, then the coordinates of the third vertex of that triangle are

• A. $(4,7)$
• B. $\left(-2,\,-\dfrac{7}{2}\right)$
• C. $(-4,-7)$
• D. $(-2,3)$

Hint:

If the orthocentre is at the origin, position vectors of vertices satisfy perpendicularity conditions using dot products or slopes.

Answer 1

The Correct Option is C

Step 1: Let the vertices of the triangle be \[ A(5,-1),\quad B(-2,3),\quad C(x,y) \] Given that the orthocentre is the origin $O(0,0)$.
Step 2: If the orthocentre is at the origin, then:

$OA \perp BC$
$OB \perp AC$

Step 3: Slope of $OA$: \[ m_{OA}=\frac{-1-0}{5-0}=-\frac{1}{5} \] Hence slope of $BC$ is: \[ m_{BC}=5 \] Equation of $BC$ through $B(-2,3)$: \[ y-3=5(x+2) \Rightarrow y=5x+13 \] Since $C(x,y)$ lies on $BC$: \[ y=5x+13 \quad \cdots (1) \]
Step 4: Slope of $OB$: \[ m_{OB}=\frac{3-0}{-2-0}=-\frac{3}{2} \] Hence slope of $AC$ is: \[ m_{AC}=\frac{2}{3} \] Equation of $AC$ through $A(5,-1)$: \[ y+1=\frac{2}{3}(x-5) \Rightarrow y=\frac{2}{3}x-\frac{13}{3} \] Since $C(x,y)$ lies on $AC$: \[ y=\frac{2}{3}x-\frac{13}{3} \quad \cdots (2) \]
Step 5: Solve equations (1) and (2): \[ 5x+13=\frac{2}{3}x-\frac{13}{3} \] Multiply by 3: \[ 15x+39=2x-13 \Rightarrow 13x=-52 \Rightarrow x=-4 \] Substitute $x=-4$ in (1): \[ y=5(-4)+13=-20+13=-7 \]
Step 6: Hence, the coordinates of the third vertex are: \[ (-4,-7) \]