Answer:
The orthocenter of the triangle is \((\frac{5}{4},1)\) and option b is correct.
Step-by-step explanation:
The ends of the base of an isosceles triangle are at (2,0) & (0,1).
Orthocentre of the triangle is the intersection point of all three altitudes from its vertices.
First find any two altitudes, then find intersection point of altitudes.
The equation of a side is x=2, which is an vertical line. The opposite vertices of this side is (0,1), so the perpendicular line is a horizontal line. The equation of first altitude is
\(y=1\) .... (1)
In an isosceles triangle, then attitude is the median of non-equal side. It means the altitude is passing through the midpoint of base side.
\(Midpoint=(\frac{2+0}{2},\frac{0+1}{2})=(1,0.5)\)
Slope of base side is
\(m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{1-0}{0-2}=-0.5\)
The product of slopes of two perpendicular lines is -1.
\(m_1\times m_2=-1\)
\(-0.5\times m_2=-1\)
\(m_2=2\)
The point slope form of a line is
\(y-y_1=m(x-x_1)\)
Where, m is slope.
The slope of second altitude is 2 and it passing through the pint (1,0.5), therefore the equation of second altitude is
\(y-0.5=2(x-1)\)
\(y=2x-1.5\) ..... (2)
From equation (1) and (2), we get
\(x=\frac{5}{4}\)
\(y=1\)
Therefore the orthocenter of the triangle is \((\frac{5}{4},1)\) and option b is correct.
