Answer:
- 7. (0, 0) - on the triangle
- 8. (1, -5) - outside the triangle
Step-by-step explanation:
Question 7
Given points:
- J(1, 3), K(-3, 1), L(0,0)
Orthocenter is the intersection of altitudes. Altitude is perpendicular segment with one end on the opposite vertices.
Lets work out two altitudes and find their intersection.
Slope of JL
- m(JL) = (0 - 3)/(0 - 1) = 3
Equation of the line that is perpendicular to JL and passes through K(-3, 1). The line has the slope -1/3.
- y - 1 = -1/3(x - (-3)) ⇒ y = -1/3x
Slope of KL
- m(KL) = (0-1)/(0 - (-3)) = -1/3
Equation of the lines that is perpendicular to KL and passes through J(1, 3). The line has a slope -1/(-1/3) = 3
- y - 3 = 3(x - 1) ⇒ y = 3x
The orthocenter is the intersection of the lines:
- -1/3x = 3x ⇒ x = 0 ⇒ y = 0
The point is (0, 0) which overlaps with point L so it is on the triangle
Question 8
Given points:
- D(-3, -2), E(-2, -2), F(1,2)
Lets work out two altitudes and find their intersection.
Slope of DE
- m(DE) = (-2 - (-2))/(-2 -(-3)) = 0, horizontal line
Equation of the line that is perpendicular to JL and passes through F(1, 2). The line is vertical:
Slope of EF
- m(EF) = (2- (-2))/(1 - (-2)) = 4/3
Equation of the lines that is perpendicular to EF and passes through D(-3,-2)
The line has a slope -1/(4/3) = -3/4:
- y - (-2) = -3/4(x - (-3)) ⇒ y = -3/4x - 17/4
The orthocenter is the intersection of the lines:
- x = 1
- y = -3/4 - 17/4 = -20/4 = -5
The point is (1, -5) is outside the triangle
