Question

The line joining two points A(2,0),B(3,1) is rotated about A in anti clockwise direction through an angle 15° . If B goes to C then C=

Topic :- Co-ordinate Geometry(Change of axes)

Answer 1

Given :-

• The line joining two points A(2,0),B(3,1) is rotated about A in anti clockwise direction through an angle 15°

• B take the position C

To Find :-

• Coordinates of C

Formula Used :-

Slope of a line

• Let us consider a line segment joining the points A (a, b) and B (c, d), then slope of line AB is represented by m and is given by

$\rm :\longmapsto\:m \: = \dfrac{d - b}{c - a}$

Solution :-

Given that

• Coordinates of A (2, 0)

• Coordinates of B (3, 1)

Let slope of AB be m and it makes an angle p with x - axis.

$\rm :\longmapsto\:m = \dfrac{1 - 0}{3 - 2}$

$\rm :\longmapsto\:m = 1$

$\rm :\longmapsto\:tanp = 1$

$\bf\implies \:p = 45 \degree$

It implies, AB makes an angle of 45° with x - axis.

Now, AB is rotated in anti-clockwise direction by an angle of 15°,

So, AC makes an angle of 60° with positive direction of x - axis,

Now,

We know, equation of line passes through (a, b) and having slope m is given by

$\rm :\longmapsto\:y - b \: = \: m(x - a)$

So, equation of line AC passes through (2, 0) and makes an angle of 60° with positive direction of x- axis is

$\rm :\longmapsto\:y - 0 = tan60 \degree \: (x - 2)$

$\bf\implies \:y = \sqrt{3}(x - 2)$

Now, we have to find coordinates of C.

Let coordinates of C be (x, y).

So,

$\rm :\longmapsto\:Coordinates \: of \: C = \bigg(x, \sqrt{3}(x - 2) \bigg)$

As AB is rotated to get new position AC

It means, Length of AB = Length of AC

$\rm :\longmapsto\:AB = AC$

On squaring both sides, we get

$\rm :\longmapsto\:AB ^{2} = AC^{2}$

Using Distance Formula, we have

$\rm :\longmapsto\: {(3 - 2)}^{2} + (1 - 0) ^{2} = {(x - 2)}^{2} + {( \sqrt{3}(x - 2)) }^{2}$

$\rm :\longmapsto\:1 + 1 = {(x - 2)}^{2} + 3 {(x - 2)}^{2}$

$\rm :\longmapsto\:2 = 4 {(x - 2)}^{2}$

$\rm :\longmapsto\: {(x - 2)}^{2} = \dfrac{1}{2}$

$\rm :\longmapsto\:x - 2 = \pm \: \dfrac{1}{ \sqrt{2} }$

$\rm :\implies\:x = 2 \: \pm \: \dfrac{1}{ \sqrt{2} }$

$\rm :\implies\:x = 2 \: + \: \dfrac{1}{ \sqrt{2} }$

and

$\rm :\implies\:x = 2 \: - \: \dfrac{1}{ \sqrt{2} } \: rejected \: as \: it \: rotates \: anticlockwise$

Now,

When

$\bf :\longmapsto\:x = 2 \: + \: \dfrac{1}{ \sqrt{2} }$

then,

$\rm :\longmapsto\:y = \sqrt{3}(2 + \dfrac{1}{ \sqrt{2}} - 2)$

$\bf :\longmapsto\:y = \sqrt{\dfrac{3}{2} }$

Hence,

$\rm :\longmapsto\:Coordinates \: of \: C = \bigg(2 + \dfrac{1}{ \sqrt{2} } , \sqrt{\dfrac{3}{2} } \bigg)$

Answer 2

Slope of AB=11

⇒tanθ=m1=1 or θ=45∘

Thus, slope of new line is tan(45∘+15∘)=tan60∘=3

(∵ it is rotated anti-clockwise, so the angle will be 45∘+15∘=60∘)

Hence, the equation is y=3x+c

But it passes through (2,0),

So, c=−23

Thus, required equation is y=3x−23

⟹ 3x−y−23=0