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).

To find the orthocenter of the triangle $ABC$ with vertices $A(1,2)$, $B(-3,4)$, and $C(5,8)$, we need to determine the intersection point of the altitudes of the triangle.

Step 1: Find the equation of the altitude from vertex A to side BC.
First, we find the slope of side $BC$. The vertices are $B(-3,4)$ and $C(5,8)$.
$m_{BC}=\frac{8-4}{5-(-3)}=\frac{4}{8}=\frac{1}{2}$
Since the altitude from $A$ (let's call it $AD$) is perpendicular to $BC$, the slope of $AD$ ($m_1$) is:
$m_1=-\frac{1}{m_{BC}}=-2$
Now, using the point-slope form for the line passing through $A(1,2)$ with slope $-2$:
$y-2=-2(x-1)$
$y-2=-2x+2$
$2x+y=4$ --- (Equation 1)

Step 2: Find the equation of the altitude from vertex B to side AC.
First, we find the slope of side $AC$. The vertices are $A(1,2)$ and $C(5,8)$.
$m_{AC}=\frac{8-2}{5-1}=\frac{6}{4}=\frac{3}{2}$
Since the altitude from $B$ (let's call it $BE$) is perpendicular to $AC$, the slope of $BE$ ($m_2$) is:
$m_2=-\frac{1}{m_{AC}}=-\frac{2}{3}$
Using the point-slope form for the line passing through $B(-3,4)$ with slope $-\frac{2}{3}$:
$y-4=-\frac{2}{3}(x-(-3\left)\right)$
$3(y-4)=-2(x+3)$
$3y-12=-2x-6$
$2x+3y=6$ --- (Equation 2)

Step 3: Solve Equation 1 and Equation 2 simultaneously to find the orthocenter.
From Equation 1, we have:
$y=4-2x$
Substitute this value of $y$ into Equation 2:
$2x+3(4-2x)=6$
$2x+12-6x=6$
$-4x=6-12$
$-4x=-6$
$x=\frac{-6}{-4}=\frac{3}{2}$
Now substitute $x=\frac{3}{2}$ back to find $y$:
$y=4-2\left(\frac{3}{2}\right)=4-3=1$

Therefore, the orthocenter of the triangle is $(\frac{3}{2},1)$.