Question

orthocenter of the triangle with vertices
(1,1),(3,5),(3,0)?

Answer 1

Given, the vertices of the triangle,

A = (1, 2)

B = (2, 6)

C = (3, -4)

Slope of AB

= y2−y1x2−x1

= 6−22–1

= 41 = 4

Slope of CF

= Perpendicular slope of AB

= −1SlopeofAB

= −14

The equation of CF is given as,

y – y1 = m(x – x1)

y + 4 = −14(x – 3)

4y + 16 = -x + 3

x + 4y = -13 ——————————– (1)

Slope of BC

= y2–y1x2–x1

= −4–63–2

= −101 = -10

Slope of AD

Perpendicular slope of BC

= −1SlopeofBC

= −1−10

= 110

The equation of AD is given as,

y – y1 = m(x – x1)

y + 2 = 110(x – 1)

10y + 20 = x – 1

x – 10y = 21 ——————————– (2)

Subtracting equation (1) and (2),

x + 4y = -13

-x + 10y = -21

——————

14y = -34

y = -2.429

Substituting the value of y in equation (1),

x + 4y = -13

x + 4(-2.429) = -13

x – 9.714 = -13

x = -3.286

Orthocenter = (-3.286,-2.429)