If ends of base of an isosceles triangle are at \( (2,0) \) and \( (0,1) \) and the equation of one side is \( x=2 \) , then the orthocentre of the triangle is
Form the figure,
\(2^2+(y_1−1)^2=y{_1}^2 \)
\(4+y_1^2+1−2y^1=y_1^2 \)
\( 5=2y_1 \) or \( y_1=5/2 \)
Equation of the line form (2,5/2)
to the given base is
\( y−5/2=2(x−2) \)
or \(2y−5=4(x−2) \)
At \(y=1\), \(−3/4=x−2\)
Or x=5/4
Reinforce your conceptual grasp with related questions from MAH MCA CET 2024:
Take timed full-length mock tests with authentic question distributions, negative marking (+12/-3 in Math), and instant All India Rank calculation.
Start Free Mock Test Series →