Question

Triangle ABC has vertices A(-5, -2), B(7, -5), and C(3, 1). Find the coordinates of the intersection of the three altitudes

Answer 1

Answer:

The coordinates of the intersection of the three altitudes = (-3.5, -1)

Step-by-step explanation:

The altitude of a triangle is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

There are therefore three altitudes possible in a triangle, one from each vertex. All three altitudes always intersect at the same point called the orthocenter of the triangle.

Let the triangle ABC have altitudes AD, BE and CF as shown in the attached image to this solution. Let the orthocentre be O.

The point O is the point where all the coordinates AD, BE and CF meet.

Hence, to obtain the coordinates of O, we just need to equate the equations of two of the lines that serve as the altitude.

Before that, we need to c9mpute the equations of the two altitudes that we will use.

Noting that the altitudes are perpendicular to the sides of the triangle, we can compute the slopes of the altitudes from caldilating the slopes of the sides.

Slope of AB

= (y₂-y₁)/(x₂−x₁)

= (-5 - (-2))/(7 - (-5))

= (-3/12)

= (-1/4)

Slope of its altitude, CF

= -1 ÷ (Slope of AB)

= -1 × (-1/4)

= 4

The equation of CF is given using point C as,

y – y₁ = m(x – x₁)

y - 1 = 4 (x – 3)

y - 1 = 4x - 12

y = 4x + 13

Slope of BC

= (y₂-y₁)/(x₂−x₁)

= (1 - (-5))/(3 - 7)

= (6/-4)

= (-3/2)

Slope of AD

= −1 ÷ (Slope of BC)

= -1 ÷ (-3/2)

= (2/3)

The equation of AD using point A given as,

y – y₁ = m(x – x₁)

y – (-2)) = (2/3) (x – (-5))

y + 2 = (2x/3) + (10/3)

y = (2x/3) + (4/3)

Now equation the equations of the altitudes CF and AD

y = 4x + 13

y = (2x/3) + (4/3)

4x + 13 = (2x/3) + (4/3)

4x - (2x/3) = (4/3) - 13

(10x/3) = (-35/3)

10x = -35

x = -3.5

y = 4x + 13

y = (4×-3.5) + 13 = -14 + 13 = -1

coordinates of the orthocentre of the triangle = (-3.5, -1)

Hope this Helps!!!