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irina [24]
3 years ago
8

100 points! Someone please help me with this! I will mark brainliest! Picture is attached below. Random answers will be reported

.

Mathematics
2 answers:
Monica [59]3 years ago
5 0

Answer:

  • 7. (0, 0) - on the triangle
  • 8. (1, -5) - outside the triangle

Step-by-step explanation:

<h3>Question 7</h3>

<u>Given points:</u>

  • 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.

<u>Slope of JL</u>

  • m(JL) = (0 - 3)/(0 - 1) = 3

Equation of the line that is perpendicular to JL and passes through K(-3, 1). <u>The line has the slope -1/3.</u>

  • y - 1 = -1/3(x - (-3)) ⇒ y = -1/3x

<u>Slope of KL</u>

  • m(KL) = (0-1)/(0 - (-3)) = -1/3

Equation of the lines that is perpendicular to KL and passes through J(1, 3). <u>The line has a slope -1/(-1/3) = 3</u>

  • y - 3 = 3(x - 1) ⇒ y = 3x

<u>The orthocenter is the intersection of the lines:</u>

  • -1/3x = 3x ⇒ x = 0 ⇒ y = 0

The point is (0, 0) which overlaps with point L so it is on the triangle

<h3>Question 8</h3>

<u>Given points:</u>

  • D(-3, -2), E(-2, -2), F(1,2)

Lets work out two altitudes and find their intersection.

<u>Slope of DE</u>

  • m(DE) = (-2 - (-2))/(-2 -(-3)) = 0, horizontal line

Equation of the line that is perpendicular to JL and passes through F(1, 2). <u>The line is vertical:</u>

  • x = 1

<u>Slope of EF</u>

  • m(EF) = (2- (-2))/(1 - (-2)) = 4/3

Equation of the lines that is perpendicular to EF and passes through D(-3,-2)

<u>The line has a slope -1/(4/3) = -3/4:</u>

  • y - (-2) = -3/4(x - (-3)) ⇒ y = -3/4x - 17/4

<u>The orthocenter is the intersection of the lines:</u>

  • x = 1
  • y = -3/4 - 17/4 = -20/4 = -5

The point is (1, -5) is outside the triangle

Vladimir [108]3 years ago
3 0

Answer:

7. Orthocenter is on the triangle, at (0,0).

8. Orthocenter is outside the triangle, at (1,-5)

Step-by-step explanation:

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