# Q.If the equations of the three sides of the triangle are x+y=1,3x+5y=2 and x-y=0 then the orthocentre of the triangle lies on the line: A) 5x - 3y=2 B) 3x -5y + 1=0 C) 2x -3y = 1 D) 5x - 3y=1 Ans:...

Q.If the equations of the three sides of the triangle are x+y=1,3x+5y=2 and x-y=0 then the orthocentre of the triangle lies on the line:

A) 5x - 3y=2

B) 3x -5y + 1=0

C) 2x -3y = 1

D) 5x - 3y=1

Ans: B.D but how???

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Let the three sides of the triangle be AB, BC and CA, so that

The equation of AB is x+y=1

of BC is 3x+5y=2

and of CA is x-y=0 (help yourself with a diagram).

Solving the equations of AB and BC, we get the coordinates of point B, which is (3/2, -1/2)

Similarly, solving the equations of BC and CA, we get the coordinates of point C, which is (1/4, 1/4), and solving the equations of CA and AB, we get the coordinates of point A, which is (1/2, 1/2).

Orthocentre of a triangle is the point where the three altitudes of the triangle intersect. Altitude is the perpendicular from a vertex on the opposite side. Let the three altitudes be AL, BM and CO.

To find the equation of AL:

AL is perpendicular on BC. Equation of BC is 3x+5y=2

`rArr` 5y=-3x+2

`rArr` y = -3/5x +2/5; its slope is -3/5

The slope of its perpendicular line = -1/(-3/5)=5/3 (`m_1m_2=-1` for two perpendicular lines)

And the equation of AL: y = 5/3x+c

Again, AL passes through A(1/2, 1/2)

So, 1/2=5/3*1/2+c

`rArr` c=-1/3

Therefore, equation of AL is y = 5/3x-1/3

`rArr` 5x-3y=1 ----(i)

Proceeding on similar lines, it can be found that the equation of BM is x+y =1 ----(ii)

and, the equation of CO is x-y=0 -----(iii)

The orthocentre lies on any or all of these three lines.

**Therefore, from eqn(i), it follows that your correct option is D).**