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MATHMAA

# 16th KVJ Solutions8b

Here you find the solutions of  16th KVJ Solutions8b  in detail explanation. Please feel free to send your opinion on this.

8Q)

(b) The line joining two points A(2, 0) and B(3, 1) is rotated about point A in anticlockwise direction through an angle of $$15^{\circ}$$. Find the equation of the line in the new position. IF B goes to C in new position, find the coordinates of C.

Solution :

Given A(2, 0) and B(3, 1)

Slope of AB = $$\frac{1-0}{3-2}$$

m   = 1

Line AB makes an angle $$\theta$$ then ,

m = Tan$$\theta$$

1= Tan $$\theta$$ , then $$theta$$ = $$45^{\circ}$$.

AB line makes $$45^{\circ}$$ angle with x-axis.

If  The line AB rotated about point A in anticlockwise direction through and angle of $$15^{\circ}$$ , B goes to C in new position.

AB changes to the line AC in new position.

AC makes an angle of $$45^{\circ} + 15^{\circ} = 60^{\circ}$$  with x-axis .

AC= AB = $$\sqrt {(3-2)^{2}+(1-0)^{2}}$$

AC = $$\sqrt{2}$$

x-coordinate of C :

x-2 = AC * Cos $$60^{\circ}$$

x = 2 + $$\frac{\sqrt{2}}{2}$$

y-coordinate of C :

y- 0 = AC * Sin $$60^{\circ}$$

y = $$\sqrt{2} *\frac{\sqrt{3}}{2}$$

y = $$\frac{\sqrt{6}}{2}$$

Hence new coordinates of C ( 2 + $$\frac{\sqrt{2}}{2}$$ ,  $$\frac{\sqrt{6}}{2}$$ )

Now equation of line AC is :

slope m= tan(\60^{\circ}\) = $$\sqrt{3}$$

Equation of line AC with slope m =$$\sqrt{3}$$ and passing through A (2, 0) is

y = $$\sqrt{3}(x-2)$$.

Therefore equation of line AC is  y = $$\sqrt{3}(x-2)$$ and coordinates of C is ( 2 + $$\frac{\sqrt{2}}{2}$$ ,  $$\frac{\sqrt{6}}{2}$$ ) .

This is the explanation of  16th KVJ Solutions8b . Please feel free to share your view on this.

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