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Please help me in getting the above problem solved,👆👆👆✍️

Please help me in getting the above problem solved,👆👆👆✍️

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Grade:11

2 Answers

Samyak Jain
333 Points
5 years ago
I.​ Reflection of (4,1) about line y = x i.e. x – y = 0 is its image about that line. Let the image be (x1,y1).So, (x1 – 4)/1 = (y1 – 1)/(-1) = –2(1.4 + (-1).1)/12 + (-1)2\therefore x1 – 4 = – (y1 – 1) = – 3x1 = – 3 + 4 = 1 & y1 = 3 + 1 = 4 [Note that coordinates of image are interchanged on reflection about y = x.You can remember this as a result.]II. The point (1,4) is shifted by 2 units towards +ve x-axis. New coordinates are (3,4).III. Distance between (3,4) and origin is r = 5 units and slope of line through (3,4) and origin is 4/3.It is rotated through an angle \Pi/4 about origin in anti-clockwise direction.Let the new slope of line be m or tan\phi.Using tan\Theta = |(m1 – m2)/(1 + m1m2)|, we get tan\Pi/4 = (m – 4/3)/(1 + 4m/3) \Rightarrow 1 = (3m – 4)/(3 + 4m)3m – 4 = 3 + 4m \Rightarrow m = – 7 = tan\phi ….(1)We infer from the sign of m that the required point lies in second quadranti.e. sin\phi > 0 & cos\phi
Samyak Jain
333 Points
5 years ago
I.​ Reflection of (4,1) about line y = x i.e. x – y = 0 is its image about that line.
 Let the image be (x1,y1).
So, (x1 – 4)/1 = (y1 – 1)/(-1) = –2(1.4 + (-1).1)/12 + (-1)2
\therefore x1 – 4 = – (y1 – 1) = – 3
x1 = – 3 + 4 = 1  &  y1 = 3 + 1 = 4  [Note that coordinates of image are interchanged on reflection about y = x.
You can remember this as a result.]
II. The point (1,4) is shifted by 2 units towards +ve x-axis.
 New coordinates are (3,4).
III. Distance between (3,4) and origin is r = 5 units and slope of line through (3,4) and origin is 4/3.
It is rotated through an angle \Pi/4 about origin in anti-clockwise direction.
Let the new slope of line be m or tan\phi.
Using tan\Theta = |(m1 – m2)/(1 + m1m2)|, we get 
tan\Pi/4 = (m – 4/3)/(1 + 4m/3)  \Rightarrow 1 = (3m – 4)/(3 + 4m)
3m – 4 = 3 + 4m  \Rightarrow m = – 7 = tan\phi            ….(1)
We infer from the sign of m that the required point lies in second quadrant
i.e. sin\phi > 0  &  cos\phi 
From (1), sin\phi = 7/\sqrt{7^{2} + 1^{2}} = 7/\sqrt{50} = 7/5\sqrt{2}
cos\phi = –1/\sqrt{7^{2} + 1^{2}} = –1/\sqrt{50} = –1/5\sqrt{2}
So, the required point is (rcos\phi,rsin\phi) \equiv (5.(-1)/5\sqrt{2} , 5.7/5\sqrt{2}) \equiv (–1/\sqrt{2} , 7/\sqrt{2})

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