Question

If the ends of the base of an isosceles triangle are at (2,0) & (0,1) & the eqn of one side is

x=2,
then the orthocentre of the triangle is

a) (3/2, 3/2)
b) (5/4, 1)
c) (3/4, 1)
d) (4/3, 7/12)
...?

Answer 1

if O is the orthocenter, O=(xo, yo)
such that xo=(2+0 + 2)/3, so xo= 4/3, so the answer must be

d) (4/3, 7/12)

Answer 2

The orthocenter of the triangle is
`((5)/(4),1)` and option b is correct.

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=((2+0)/(2),(0+1)/(2))=(1,0.5)`

Slope of base side is

`m_1=(y_2-y_1)/(x_2-x_1)=(1-0)/(0-2)=-0.5`

The product of slopes of two perpendicular lines is -1.

`m_1* m_2=-1`

`-0.5* 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=(5)/(4)`

`y=1`

Therefore the orthocenter of the triangle is
`((5)/(4),1)` and option b is correct.