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Q. The vertices of a triangle in the argand plane are `3+4i, 4+3i` and `2sqrt6+i,` then...

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user8235304 | Student, Grade 11 | (Level 1) Valedictorian

Posted August 20, 2013 at 3:35 AM via web

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Q. The vertices of a triangle in the argand plane are `3+4i, 4+3i` and `2sqrt6+i,` then distance between orthocentre and circumcentre of the triangle is equal to

A) `sqrt(137-28sqrt6)`

B) `sqrt(137+28sqrt6)`

C) `(1/2)sqrt(137+28sqrt6)`

D) `(1/3)sqrt(137+28sqrt6)`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 20, 2013 at 2:17 PM (Answer #1)

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The coordinates of vertices of triangle, in Argand plane, are `A(3,4), B(4,3), C(2sqrt6,1)` .

You need to remember that the orthocenter of triangle is the point of intersection of the three altitudes of triangle.

You need to find the point slope form of equation of altitude, using the relation between the slopes of two perpendicular lines, such that:

`AD _|_ BC => m_(AD)*m_(BC) = -1 => m_(AD) = -1/(m_(BC))`

`y - y_A = m_(AD)(x - x_A)`

You need to evaluate `m_(BC)` using the slope formula, such that:

`m_(BC) = (y_B - y_C)/(x_B - x_C) => m_(BC) = (3 - 1)/(4 - 2sqrt6) => m_(BC) = 2/(2(2 - sqrt6)) => m_(BC) = 1/(2 - sqrt6)`

`m_(AD) = -1/(1/(2 - sqrt6)) => m_(AD) = sqrt6 - 2`

`y - 4 = (sqrt6 - 2)(x - 3) => y = (sqrt6 - 2)(x - 3) + 4`

`y - y_B = m_(BE)(x - x_B)`

`m_(BE) = -1/(m_(AC))`

`m_(AC) = (1-4)/(2sqrt6 - 3) => m_(AC) = -3/(2sqrt6 - 3)`

`m_(BE) = (2sqrt6 - 3)/3`

`y - 3 = (2sqrt6 - 3)/3(x - 4) => y = (2sqrt6 - 3)/3(x - 4) + 3`

You need to solve the following system of equations to evaluate to evaluate the orthocenter Q, such that:

`{(y = (sqrt6 - 2)(x - 3) + 4),(y = (2sqrt6 - 3)/3(x - 4) + 3):} => (sqrt6 - 2)(x - 3) + 4 = (2sqrt6 - 3)/3(x - 4) + 3`

`sqrt6*x - 3sqrt6 - 2x + 6 + 4 = 2sqrt6/3*x - 8sqrt6/3 - x + 4 + 3`

`sqrt6*x- 2x - 2sqrt6/3*x + x = -10 + 3sqrt6 + 7 - 8sqrt6/3 `

`sqrt6/3*x - x = -3 + sqrt6/3`

`x(sqrt6/3 - 1) = sqrt6/3 - 3 => x = (sqrt6 - 9)/(sqrt6 - 3)`

`y = sqrt6*((sqrt6 - 9)/(sqrt6 - 3)) - 3sqrt6 - 2*((sqrt6 - 9)/(sqrt6 - 3)) + 6 + 4`

`y = (6 - 9sqrt6 - 3 + 9sqrt6 - 2sqrt6 + 18 + 10sqrt6 - 30)/(sqrt6 - 3)`

`y = (-9 + 8sqrt6)/(sqrt6 - 3)`

=> Q  `((sqrt6 - 9)/(sqrt6 - 3),(-9 + 8sqrt6)/(sqrt6 - 3))`

You need to evaluate the coordinates of circumcenter C that is the point of intersection of perpendicular bisectors of triangle.

You first need to fiind the midpoints of the sides BC and AC, such that:

`x_(M) = (x_B + x_C)/2 => x_M = (2 + sqrt6)`

`y_M = (y_B + y_C)/=> y_M = 2`

`x_N = (x_A + x_C)/2 => x_N = (3 + 2sqrt6)/2`

`y_N = 5/2`

`y - y_M = m_(CM)(x - x_M)`

`m_(CM) = -1/(m_(BC)) => m_(CM) = sqrt6 - 2`

`y - 2 = (sqrt6 - 2)(x - 2 - sqrt6) => y = (sqrt6 - 2)(x - (sqrt6 + 2)) + 2`

`y = (sqrt6 - 2)x - (6 - 4) + 2`

`y = (sqrt6 - 2)x`

`y - y_N = m_(CN)(x - x_N)`

`y - 5/2 = ((2sqrt6 - 3)/3)(x - (3 + 2sqrt6)/2)`

`y = ((2sqrt6 - 3)/3)*x - (24 - 9)/6 + 5/2`

`y = ((2sqrt6 - 3)/3)*x - 15/6 + 5/2`

`y = ((2sqrt6 - 3)/3)*x - 5/2 + 5/3`

`y = ((2sqrt6 - 3)/3)*x`

You need to solve the following system of equations to evaluate to evaluate the orthocenter C, such that:

`{(y = (sqrt6 - 2)x),( ((2sqrt6 - 3)/3)*x):} => (sqrt6 - 2)x = ((2sqrt6 - 3)/3)*x`

`(sqrt6 - 2)x - ((2sqrt6 - 3)/3)*x = 0 => x(sqrt6 - 2 -(2sqrt6 - 3)/3) = 0 => x = 0 => y = 0`

=> C (0,0

You need to evaluate the distance between the circumcenter and orthocenter, using distance formula, such that:

`[QC] = sqrt((x_Q - x_C)^2 + (y_Q - y_C)^2)`

`[QC] = sqrt(((sqrt6 - 9)/(sqrt6 - 3))^2 + ((-9 + 8sqrt6)/(sqrt6 - 3))^2)`

`[QC] = (sqrt(6 + 81 - 18sqrt6 + 81 + 384 - 144sqrt6))/|(sqrt6 - 3)|`

`[QC] = (sqrt(6 + 81 - 18sqrt6 + 81 + 384 - 144sqrt6))/|(sqrt6 - 3)|`

`[QC] = (sqrt(552 - 162sqrt6))/|(sqrt6 - 3)|`

`[QC] = ((sqrt6 + 3)sqrt(552 - 162sqrt6))/|(6 - 9)|`

`[QC] = (1/3)(((sqrt6 + 3)sqrt(552 - 162sqrt6))) = (1/3)sqrt(28sqrt6 + 137)`

Hence, evaluating the distance between orthocenter and circumcenter, yields the answer` D) (1/3)sqrt(28sqrt6 + 137).`

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