Question

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

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

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.