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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 4×2511−7×3011+k=0⇒k=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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