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)

The orthocenter of the triangle is 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

                             .... (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.

Slope of base side is

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

The point slope form of a line is

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

               ..... (2)

From equation (1) and (2), we get

Therefore the orthocenter of the triangle is and option b is correct.

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2