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Analytical Geometry

Scorching goIITian

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9 Apr 2009 21:13:23 IST

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the equation to the sides of a triangle are x-3y =0 ,, 4x+3y=5 and 3x+y =0. the line 3x-4y pases thr

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the equation to the sides of a triangle are x-3y =0 ,, 4x+3y=5 and 3x+y =0. the line 3x-4y pases through..A)the incentrre b) the orthocentre c)the centroid d) the circumcentre

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edison

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13 Aug 2010 16:28:54 IST

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Find the vertexof the triangle by finding intersection points on three lines

Here simultaneous solution of x-3y =0 and 4x+3y=5 gives say vertex 'A'

simultaneous solutionx-3y =0 and 3x+y =0 gives vertex 'B'

and simultaneous solution 4x+3y=5 and 3x+y =0 gives 'C'

Now find the centroid by formula

X_c = (x₁ + x₂ + x₃) / 3

and Y_c = (y₁+ y₂+y₃ ) / 3

Check whether (X_c, Y_c) is a solution of 3x-4y = 0, if yes then it passes the centroid else not.

Rajdeep chakraborty

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6 Sep 2010 17:36:11 IST

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Remember that the orthocentre of a right angled triangle is the vertex containing the right angle. Here u can see that the lines x=3y and y= - 3x are mutually perpendicular bcoz the product of their slopes is = (1/3).(-3) = -1 .

Therefore the three lines form a right angled triangle.

Both the lines, x=3y and y= -3x pass through the origin and hence their intersection point is the origin. Therefore the orthocentre of the triangle formed by the given st. lines is the origin. Now , the st. line 3x-4y=0 or 3x=4y also passes through the origin and hence it passes through the orthocentre of the given triangle .

You may also keep it in mind that the circumcentre of a right angled triangle is the mid point of the hypotenuse .

Yagyadutt Mishra

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6 Sep 2010 19:43:56 IST

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Question is incomplete ....i guess ?

Ques: the equation to the sides of a triangle are x-3y =0 ,, 4x+3y=5 and 3x+y =0. the line 3x-4y =? pases through..A)the incentrre b) the orthocentre c)the centroid d) the circumcentre

Incentre : Point where all the angle bisector intersect

Orthocentre: Point where all the altitude of the triangle intersect.

Circumcentre: Point where all the perpendicular bisector of the triangle intersect.

Centroid : The point where all the medians intersect ....

Three sides x - 3y = 0 4x +3y = 5 3x + y = 0

In this case the two lines are perpendicular to each other....that is 3x + y = 0 and x-3y = 0 ...

Hence the triangle is a right angle triangle with hypotenous 4x + 3y = 5

Now in question you have given the line to which we have to choose the option is

3x-4y = K(i suppose)

Now 3x-4y = K is perpendicular to hypotenous .......since it is perpendicular ....so it can pass through orthocentre.....or circumcentre....

Answer will depend on the value of K....

How ?

You will solve x-3y = 0 and 3x +y = 0 ...i.e (0,0) ...if (0,0) satisfy the 3x - 4y = k then it will be orthocentre.....otherwise orthocentre is not the correct option...

Also ...if it will be circumcentre only if ...it passes through middle point of hypotenous.....

Hence K should be given in the question ..definately

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