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if origin is the orthocentre ∆ABC where A(5,-1),B(-2,3) then the orthocentre of ∆OAC is

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9 months ago

```							Orthocentre is the point of intersection of all altitudes of a triangle Let A=(x1,y1) B=(5,-1) and C=(-2,3) be the vertices of triangle Orthocentre, O=(0,0) Slope of BC = -4/7 Altitude through A and perpendicular to BC is (y-y1) = 7/4(x-x1) --> 4y-4y1 = 7x-7x1 this passes through orthocentre i.e., (0,0) therefore 4y1 = 7x1 ........................................... Slope of AC= (3-y1)/(-2-x1) Now, altitude through B and perpendicular to AC is y+1 = (x1+2)/(3-y1) [y-5] this passes through (0,0) therefore 5x1-y1+13=0 ....................................(2) On solving (1) and (2) we get (x1,y1) = (-4,-7) So A=(-4,-7) [OR] 2)EASY PROCESS: Find orthocentre of given points(i.e., O,B,C) that is the orthocenter of triangle ABC Important Property: "If O is orthocentre of triangle ABC then the four points O,A,B,C are such that each point is orthocentre of triangle formed by other three points" Hint : If one of the vertices of a triangle is origin, with remaining vertices (x1,y1) and (x2,y2), then orthocentre is ( k{y2-y1} , k{x1-x2} ) where k=(x1x2 + y1y2) / (x1y2 - x2y1)
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9 months ago
```							Dear student Ans foe the above question will be (-4,-7) , It is not mentioned in the above .Good Luck Cheers
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9 months ago
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