Question

Perpendicular to the line joining the points A(1.2), B(5,6).
The slope of a straight line through A(3,2) is 3/4. Find the co-ordinates of the points on the line that are 5 units away from it.

Answer 1

Given a point $A(3,2)$ on a line with slope $\frac{3}{4}$

Equation of the line will be :

$\Rightarrow y-2=\frac{3}{4}(x-3)$

$\Rightarrow 4y-8=3x-9$

$\Rightarrow 3x-4y-1=0$

We are supposed to find a point on the above line which is 5 units away from $A(3,2).$

We know that the slope of a line is $m=\tan \theta$ where, $\theta$ is basically the angle that the line makes with the x-axis.

( Diagram : 1 )

Now the coordinates of a point on the line with slope $\tan \theta$ at a distance ${}^{\prime }{r}^{\prime }$ units from $(h,k)$ which is also on the line is given by

$\Rightarrow x=h\pm r\cos \theta$

$y=k\pm r\sin \theta$

${}^{\prime }{\pm }^{\prime }$ is used since we could have a point ${}^{\prime }{r}^{\prime }$ units from $(h,k)$ in either direction on the line.

Using the above concept, we have

( Diagram : 2 )

$\Rightarrow \tan \theta =\frac{3}{4}$

$\Rightarrow \cos \theta =\frac{4}{5}$

$\Rightarrow \sin \theta =\frac{3}{5}$

$\therefore x=3\pm 5\cos \theta =3\pm 5\times \frac{4}{5}=3\pm 4=7or-1$

$y=4\pm 5\sin \theta =4\pm 5\times \frac{3}{5}=4\pm 3=7or1$

$\therefore$ the coordinates that are 5 units away from $A(3,2)$ on the line of slope $\frac{3}{4}$ are $(7,7,)$ and $(-1,1)$

Hence, solved.