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Let k be an integer such that the triang...

Let `k` be an integer such that the triangle with vertices `(k ,-3k),(5, k)` and `(-k ,2)` has area `28s qdot` units. Then the orthocentre of this triangle is at the point : `(1,-3/4)` (2) `(2,1/2)` (3) `(2,-1/2)` (4) `(1,3/4)`

A

`(2,(1)/(2))`

B

`(2,-(1)/(2))`

C

`(1,(3)/4)`

D

`(1,-(3)/4)`

Text Solution

Verified by Experts

The correct Answer is:
A
We have
`(1)/(2)|{:(k,,-3k,,1),(5,,k,,1),(-k,,2,,1):}|=+-28`
`rArr5k^2+13k-46=0`
or `5k^2+13k+66=0` (no ral solution)
`rArr=K=(-23)/(5)ork=2`
Since K is an integer, `k=2`

Form the figure, slope of BC is 0. So, equation of AD is
`x=2`
Slope of AC is `1//2`.
Eqaution of line BE is `x-2y-=0`
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Knowledge Check

  • The orthocentre of the triangle whose vertices are (5,-2), (-1, 2) and (1,4) is

    A
    `((1)/(5), (14)/(5))`
    B
    `((14)/(5), (1)/(5))`
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    D
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    B
    ` -2`
    C
    ` -12,-2`
    D
    `12,-2`

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