JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 1)
In a triangle PQR, the co-ordinates of the points P and Q are ($$-$$2, 4) and (4, $$-$$2) respectively. If the equation of the perpendicular bisector of PR is 2x $$-$$ y + 2 = 0, then the centre of the circumcircle of the $$\Delta$$PQR is :
($$-$$1, 0)
(1, 4)
(0, 2)
($$-$$2, $$-$$2)
Explanation
Mid point of $$PQ \equiv \left( {{{ - 2 + 4} \over 2},{{4 - 2} \over 2}} \right) \equiv (1,1)$$
Slope of $$PQ = {{4 + 2} \over { - 2 - 4}} = - 1$$
Slope of perpendicular bisector of PQ = 1
Equation of perpendicular bisector of PQ
$$y - 1 = 1(x - 1)$$
$$ \Rightarrow y = x$$
Solving with perpendicular bisector of PR,
2x $$-$$ y + 2 = 0
Circumcentre is ($$-$$2, $$-$$2)
Slope of $$PQ = {{4 + 2} \over { - 2 - 4}} = - 1$$
Slope of perpendicular bisector of PQ = 1
Equation of perpendicular bisector of PQ
$$y - 1 = 1(x - 1)$$
$$ \Rightarrow y = x$$
Solving with perpendicular bisector of PR,
2x $$-$$ y + 2 = 0
Circumcentre is ($$-$$2, $$-$$2)
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