Question

If the equations of the three sides of a 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. $5x-3y=1$

Answer 1

Answer: (B), (D)

The correct options are
B $3x-5y+1=0$
D $5x-3y=1$

Let the point of intersection of lines $x+y=1$ and $3x+5y=2$ be $A(h,k)$

By solving those two equations , we get $h=\frac{3}{2},k=-\frac{1}{2}$

Let the point of intersection of lines $x+y=1$ and $x-y=0$ be $B(p,q)$

By solving above equations , we get $p=\frac{1}{2},q=\frac{1}{2}$

The equation of line passing through $A$ and perpendicular to line $x-y=0$ is $(y+\frac{1}{2})=-1(x-\frac{3}{2})$

$\Rightarrow x+y=1$

Therefore point $B$ is orthocenter

point $B(\frac{1}{2},\frac{1}{2})$ lies on $3x-5y+1=0$ and $5x-3y=1$

Therefore the correct option is $B,D$