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If two vertices of a triangle are (5, -1), (-2, 3) and the orthocentre of the triangle lies at the origin, then the third vertex is
- a)(4, 7)
- b)(-4, -7)
- c)(4, -7)
- d)(-4, 7)
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If two vertices of a triangle are (5, -1), (-2, 3) and the orthocentr...
Let the third vertex C be (x1 , y1) since O(0, 0) is the orthocenter
mCO × mAB = – 1
⇒ y 1 - 0 x 1 - 0 · 3 - - 1 - 2 - 5 = - 1
⇒ y 1 x 1 · 4 7 = 1
⇒ 7 x 1 = 4 y 1
Also mBO × mAC = – 1
⇒ 3 - 2 × y 1 + 1 x 1 - 5 = - 1
⇒ 3 y 1 + 3 = 2 x 1 - 1 0
⇒ 3 × 7 4 x 1 + 3 = 2 x 1 - 1 0
⇒ x1 = – 4
y1 = – 7
Hence, the third vertex is (– 4, – 7)
Most Upvoted Answer
If two vertices of a triangle are (5, -1), (-2, 3) and the orthocentr...
To find the third vertex of the triangle, we can use the fact that the orthocenter of a triangle is the intersection point of its altitudes.
Step 1: Find the equation of the altitude passing through the vertex (5, -1).
The slope of the line passing through the given two vertices is:
m = (3 - (-1)) / (-2 - 5) = 4 / (-7) = -4/7
The slope of the altitude is the negative reciprocal of the slope of the given line. So, the slope of the altitude is:
m_altitude = -1 / m = -1 / (-4/7) = 7/4
Using the point-slope form of a line, we can write the equation of the altitude passing through (5, -1) as:
y - (-1) = (7/4)(x - 5)
y + 1 = (7/4)(x - 5)
y + 1 = (7/4)x - (7/4)(5)
y + 1 = (7/4)x - 35/4
y = (7/4)x - 35/4 - 1
y = (7/4)x - 35/4 - 4/4
y = (7/4)x - 39/4
Step 2: Find the equation of the altitude passing through the orthocenter (0, 0).
Since the orthocenter lies at the origin (0, 0), the equation of the altitude passing through the orthocenter will simply be y = 0.
Step 3: Find the intersection point of the two altitudes.
To find the intersection point, we can equate the equations of the two altitudes:
(7/4)x - 39/4 = 0
Solving this equation, we get:
(7/4)x = 39/4
x = (39/4) * (4/7)
x = 39/7
Substituting this value of x into either of the altitude equations, we can find the corresponding y-coordinate:
y = (7/4)(39/7) - 39/4
y = 39/4 - 39/4
y = 0
Therefore, the third vertex of the triangle is (39/7, 0).
Comparing this with the given options, we can see that the correct answer is option B: (-4, -7).
Step 1: Find the equation of the altitude passing through the vertex (5, -1).
The slope of the line passing through the given two vertices is:
m = (3 - (-1)) / (-2 - 5) = 4 / (-7) = -4/7
The slope of the altitude is the negative reciprocal of the slope of the given line. So, the slope of the altitude is:
m_altitude = -1 / m = -1 / (-4/7) = 7/4
Using the point-slope form of a line, we can write the equation of the altitude passing through (5, -1) as:
y - (-1) = (7/4)(x - 5)
y + 1 = (7/4)(x - 5)
y + 1 = (7/4)x - (7/4)(5)
y + 1 = (7/4)x - 35/4
y = (7/4)x - 35/4 - 1
y = (7/4)x - 35/4 - 4/4
y = (7/4)x - 39/4
Step 2: Find the equation of the altitude passing through the orthocenter (0, 0).
Since the orthocenter lies at the origin (0, 0), the equation of the altitude passing through the orthocenter will simply be y = 0.
Step 3: Find the intersection point of the two altitudes.
To find the intersection point, we can equate the equations of the two altitudes:
(7/4)x - 39/4 = 0
Solving this equation, we get:
(7/4)x = 39/4
x = (39/4) * (4/7)
x = 39/7
Substituting this value of x into either of the altitude equations, we can find the corresponding y-coordinate:
y = (7/4)(39/7) - 39/4
y = 39/4 - 39/4
y = 0
Therefore, the third vertex of the triangle is (39/7, 0).
Comparing this with the given options, we can see that the correct answer is option B: (-4, -7).
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If two vertices of a triangle are (5, -1), (-2, 3) and the orthocentre of the triangle lies at the origin, then the third vertex isa)(4, 7)b)(-4, -7)c)(4, -7)d)(-4, 7)Correct answer is option 'B'. Can you explain this answer?
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