JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 13)
The image of the point (3, 5) in the line x $$-$$ y + 1 = 0, lies on :
(x $$-$$ 4)2 + (y $$-$$ 4)2 = 8
(x $$-$$ 4)2 + (y $$+$$ 2)2 = 16
(x $$-$$ 2)2 + (y $$-$$ 2)2 = 12
(x $$-$$ 2)2 + (y $$-$$ 4)2 = 4
Explanation
So, let the image is (x, y)
So, we have
$${{x - 3} \over 1} = {{y - 5} \over { - 1}} = - {{2(3 - 5 + 1)} \over {1 + 1}}$$
$$ \Rightarrow $$ x = 4, y = 4
$$ \Rightarrow $$ Point (4, 4)
Which will satisfy the curve
(x $$-$$ 2)2 + (y $$-$$ 4)2 = 4
as (4 $$-$$ 2)2 + (4 $$-$$ 4)2 = 4 + 0 = 4
So, we have
$${{x - 3} \over 1} = {{y - 5} \over { - 1}} = - {{2(3 - 5 + 1)} \over {1 + 1}}$$
$$ \Rightarrow $$ x = 4, y = 4
$$ \Rightarrow $$ Point (4, 4)
Which will satisfy the curve
(x $$-$$ 2)2 + (y $$-$$ 4)2 = 4
as (4 $$-$$ 2)2 + (4 $$-$$ 4)2 = 4 + 0 = 4
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