Question

Geometry Unit 7 Lesson 7. Please Help Picture attached.

Answer 1

Answer:

The coordinates of orthocenter of the triangle are (1,5).

Step-by-step explanation:

The given vertices are A(0,6), B(4,6) and C(1,3).

Orthocenter of a triangle is the intersection point of all altitudes.

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

The slope of AB is

$m_{AB}=\frac{y_2-y_1}{x_2-x_1}=\frac{6-6}{4-0}=0$

The slope of AB is 0, therefore slope of line which is perpendicular to AB is $\frac{-1}{0}$.

Point slope form of a line is

$y-y_1=m(x-x_1)$

Where, m is the slope.

The equation of altitude on AB form C is

$y-3=\frac{1}{0}(x-1)$

$0=x-1$

$x=1$                              ..... (1)

Slope of BC is

$m_{BC}=\frac{y_2-y_1}{x_2-x_1}=\frac{6-3}{1-4}=1$

The slope of BC is 1 therefore the slope of altitude on BC from A is -1.

The equation of altitude on BC from A is

$y-6=-1(x-0)$

$y=-x+6$                         ..... (2)

Using (1) and (2) we get

$x=1, y=5$

The  intersection point of two altitudes is (1,5). Therefore coordinates of orthocenter of the triangle are (1,5).