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Question

Orthocentre of the triangle whose sides are given by 4x - 7y + 10 = 0, x + y - 5 = 0 & 7x + 4y - 15 = 0 is

A
(-1, -2)
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B
(1,- 2)
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C
(-1, 2)
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D
(1, 2)
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Solution

The correct option is D (1, 2)

Let ABC be the triangle and point O be the orthocentre of the triangle ABC.
Let equation of side AB be 4x - 7y + 10 = 0
4x - 7y =-10 . . . . (1)
side BC is x + y = 5 . . . (2)
side AC is 7x + 4y = 15 . . . (3)
Solving equation (1) and (2), we get
x=2511, y=3011
So, point B will be (2511,3011)
And,
Solving equation (2) and (3), we get
x=53, y=203
So, point C will be (53, 203)
And, Solving equation (1) and (3), we get
x = 1, y = 2
So, point A will be (1, 2)
Now, altitude AD is perpendicular to BC, therefore equation of AD is x - y + k = 0
AD is passing through the point A(1, 2) so 1 - 2 + k = 0 k = 1
Thus, equation of AD = x - y + 1 = 0 . . . (4)
And, altitude BE is perpendicular to AC, therefore equation of BE is 4x - 7y + k = 0
BE is passing through the point B(2511,3011) so 4×25117×3011+k=0k=10
Thus, equation of BE = 4x - 7y + 10 = 0 . . . (5)
Solving equation (4) and (5), we get
x = 1 , y = 2
Thus, the coordinate of the orthocentre is (1,2)
Hence option (d) is correct

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