Question

A line passes through $P(4,5)$ and makes ${30}^o$ with X-axis. Then co-ordinates of a point which is at a distance $4units$ on either side of $P$, is:

• A. $(4-2\sqrt{3},7)$`
• B. $(4\pm 2\sqrt{3},7)$
• C. $(4+2\sqrt{3},7)$, $(4-2\sqrt{3},7)$
• D. $(4+23,7)$

Answer 1

The correct option is D $(4+23,7)$

Given line makes angle ${30}^0$ with x-axis

Hence slope of line $m=\tan {30}^0=\frac{1}{\sqrt{3}}$

Equation of line from point $P(4,5)$ and slope m

$y-5=\frac{1}{\sqrt{3}}(x-4)$

$\sqrt{3}y-5\sqrt{3}=x-4$

$x-y\sqrt{3}+5\sqrt{3}-4=0$

Let a point $Q(x_1,y_1)$ is at the distance of 4 unit from Point P

By distance formula

$\sqrt{(x_1-4{)}^2+(y_1-5{)}^2}=4$---(1)

Here Point Q lie on line $x-y\sqrt{3}+5\sqrt{3}-4=0$

Hence $x_1=4+y_1\sqrt{3}-5\sqrt{3}$

From eq (1)

$\sqrt{(4+y_1\sqrt{3}-5\sqrt{3}-4{)}^2+(y_1-5{)}^2}=4$

On squaring both sides

$3y_1^2+75-30y_1+y_1^2+25-10y_1=16$

$4y_1^2-40y_1+84=0$

$y_1^2-10y_1+21=0$

$y_1=\frac{10\pm \sqrt{100-84}}{2}$

$y_1=\frac{10\pm \sqrt{16}}{2}$

$y_1=\frac{10\pm 4}{2}$

$y_1=7$ and $y_1=3$

$x_1=4+7\sqrt{3}-5\sqrt{3}$ and $x_1=4+3\sqrt{3}-5\sqrt{3}$

$x_1=4+2\sqrt{3}$ and $x_1=4-2\sqrt{3}$

$Q(4+2\sqrt{3},7)$ and $Q(4-2\sqrt{3},3)$