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  • If two sides of a triangle are 4x + 5y – 20 = 0 and 3x – 2y + 6 = 0 with orthocentre (1, 1). Then equation ofthird sides is-

If two sides of a triangle are 4x + 5y – 20 = 0 and 3x – 2y + 6 = 0 with orthocentre (1, 1). Then equation of  third sides is-

Answers (1)

best_answer

As we now

The orthocenter is the intersection of altitudes

Let Triangle be ?ABC 
In which CM is perpendicular to AB
and BN is perpendicular to AC

And here we have to find the equation of line BC

At first, we have to find altitude perpendicular to line 4x+5y-20=0 and passing through (1,1) that means we have to equation of CM
which we get CM:- 5x-4y-1=0

The same way we have to find the altitude perpendicular to the line 3x-2y+6=0
and passing through (1,1) that means we have to find the equation of BN which we get 
BN :- 2x+3y-5=0

Now we have to find the intersection point of AC and CM which we get coordinate of Point C
which is (-13,-33/2)

The same way we have to find the intersection point of AB and BN which we get coordinate of point B which is (35/2,-10)

Now as we have to find the equation of line BC
and we know point B and C i.e.
B(35/2,-10) and C(-13,-33/2)

By the two-point form of a line, we get the equation of line BC and that will be 26x-122y-1675=0

Answer:- 26x-122y-1675=0

 

Posted by

Pankaj Sanodiya

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