Let `M(2,13/8)` is the circumcentre of `DeltaPQR` whose sides `PQ and PR` are represented by the straight lines `4x-3y = 0 and 4x + y = 16` respectively. The orthocentre of `DeltaPQR` is

Options

A. orthocentre of `DeltaPQR` is `(3, (3)/(4))`
B. orthocentre of `DeltaPQR` is `((7)/(3), (4)/(3))`
C. orthocentre of `DeltaPQR` is `((4)/(3), (7)/(3))`
D. coordinates of point `R` is `(4, 0)`

Answer

A, D

Explanation

On solving `4x - 3y = 0` and `4x + y = 16`, we get `P(3,4)`.
The equation of lines perpendicular to `PQ` is `3x + 4y = k`
As above line passes through `M(2, (13)/(8))`
So `3(2) + 4((13)/(8)) = k rArr k = (25)/(2)`
`:.` Equation of `AM` is `3x + 4y = (25)/(2)`
Similarly, the equation of line perpendicular to
`PR` is `x - 4y = lambda`
As above line passes through `M(2, (13)/(8))`,
So `2 - 4((13)/(8)) = lambda rArr lambda = -(9)/(2)`
`:.` Equation of `CM` is `x - 4y = -(9)/(2)`
Now, on solving `4x - 3y = 0` and `3x + 4y = (25)/(2)`,
We get `A((3)/(2),2)`
Also on solving `4x + y = 16` and `x - 4y = -(9)/(2)`,
we get `C((7)/(2),2)`
As `A` is midpoint of `PQ`, so `Q = (0,0)` and `C` is the midpoint of `PR`, so `R = (4, 0)`

`(2, (13)/(8)) , ((7)/(3), (4)/(3))`
`:.` By applicant of section formula, we get
`(1(alpha) + 2(2))/(1 + 2) = (7)/(3)`
`rArr alpha = 7 - 4 = 3`
And `(1(beta) + 2((13)/(8).))/(1 + 2) = (4)/(3)`
`rArr beta 4 -(13)/(4) = (16 - 13)/(4) = (3)/(4)`
Hence orthocentre of `DeltaPQR` is `(3, (3)/(4))`