If two vertices of a triangle are $$(5, -1)$$ and $$(-2, 3)$$ and its orthocentre is at $$(0, 0)$$, then the third vertex is

Let the given vertices be $$A(5,-1)$$ and $$B(-2,3)$$, and let the unknown third vertex be $$C(x,y)$$. The orthocentre is $$H(0,0)$$, i.e. the common intersection point of the three altitudes of the triangle.

Step 1: Slope of altitude through A
The altitude from $$A$$ passes through $$A(5,-1)$$ and $$H(0,0)$$.
Slope of this altitude, $$m_{AH}=\dfrac{-1-0}{5-0}=-\dfrac15$$.

This altitude is perpendicular to side $$BC$$, so
$$m_{BC}\;=\;-\dfrac1{m_{AH}}=5.$$

Therefore, for points $$B(-2,3)$$ and $$C(x,y)$$, we have
$$m_{BC}=\dfrac{y-3}{\,x+2\,}=5\quad\Longrightarrow\quad y-3=5(x+2).$$ Hence

$$y = 5x + 13 \quad -(1)$$

Step 2: Slope of altitude through B
The altitude from $$B$$ passes through $$B(-2,3)$$ and $$H(0,0)$$.
Slope of this altitude, $$m_{BH}=\dfrac{3-0}{-2-0}=-\dfrac32$$.

This altitude is perpendicular to side $$AC$$, so
$$m_{AC}\;=\;-\dfrac1{m_{BH}}=\dfrac23.$$

Therefore, for points $$A(5,-1)$$ and $$C(x,y)$$, we have
$$m_{AC}=\dfrac{y+1}{\,x-5\,}=\dfrac23\quad\Longrightarrow\quad 3(y+1)=2(x-5).$$ Hence

$$3y + 3 = 2x - 10 \;\Longrightarrow\; 3y = 2x - 13 \;\Longrightarrow\; y=\dfrac{2x-13}{3}\quad -(2)$$

Step 3: Solve simultaneous equations for C(x, y)
Equate the two expressions for $$y$$ from (1) and (2):

$$5x + 13 = \dfrac{2x-13}{3}.$$

Multiply by 3:
$$15x + 39 = 2x - 13.$$

$$13x = -52 \;\Longrightarrow\; x = -4.$$

Substitute $$x=-4$$ in (1):
$$y = 5(-4) + 13 = -20 + 13 = -7.$$

Thus the third vertex is $$C(-4,-7)$$.

Option B which is: $$(-4,\,-7)$$