The vertices of a triangle ABC are A(1, 2), B(–3, 4), C(5, 8), then orthocentre of △ABC is

The correct option is (3/2, 1).

Step-by-step Explanation:
The orthocentre of a triangle is the point of intersection of its altitudes. As shown in the provided diagram, we have a triangle $△ABC$ with vertices:
$A(1, 2)$
$B(-3, 4)$
$C(5, 8)$
We will find the equations of two altitudes, $AD$ (altitude from vertex $A$ to side $BC$) and $BE$ (altitude from vertex $B$ to side $AC$), and solve them to find their intersection point, which is the orthocentre $H$.

Step 1: Find the equation of altitude $AD$
First, find the slope of side $BC$ using the slope formula:

$m_{BC}=\frac{y_2-y_1}{x_2-x_1}=\frac{8-4}{5-(-3)}=\frac{4}{8}=\frac{1}{2}$

Since the altitude $AD$ is perpendicular to $BC$, its slope ($m_{AD}$) is the negative reciprocal of $m_{BC}$:

$m_{AD}=-\frac{1}{m_{BC}}=-2$

Using the point-slope form, the equation of the line representing altitude $AD$ passing through $A(1, 2)$ is:

$y-2=-2(x-1)$

Simplifying this equation:
$y-2=-2x+2$
$2x+y=4$    — (Equation 1)

Step 2: Find the equation of altitude $BE$
Next, find the slope of side $AC$:

$m_{AC}=\frac{8-2}{5-1}=\frac{6}{4}=\frac{3}{2}$

Since the altitude $BE$ is perpendicular to $AC$, its slope ($m_{BE}$) is the negative reciprocal of $m_{AC}$:

$m_{BE}=-\frac{1}{m_{AC}}=-\frac{2}{3}$

Using the point-slope form, the equation of the line representing altitude $BE$ passing through $B(-3, 4)$ is:

$y-4=-\frac{2}{3}(x+3)$

Simplifying this equation:
$3(y-4)=-2(x+3)$
$3y-12=-2x-6$
$2x+3y=6$    — (Equation 2)

Step 3: Solve the linear equations to find the orthocentre
We have the following system of linear equations:
1) $2x+y=4$
2) $2x+3y=6$
Subtract Equation 1 from Equation 2:

$(2x+3y)-(2x+y)=6-4$
$2y=2$
$y=1$

Substitute $y=1$ back into Equation 1:
$2x+1=4$
$2x=3$

$x=\frac{3}{2}$

Thus, the intersection point of the altitudes is the orthocentre $H\left(\frac{3}{2},1\right)$.