Question

If the slope of a line passing through the point $A(3,2)$ is $\frac{3}{4}$, then which of the following are the coordinates of the point lie on the line which are 5 units away from the point $A$ ?

• A. $(-1,-1)$
• B. $(7,5)$
• C. Both(a) and (b)
• D. None of these

Answer 1

Answer: (C)

Equation of a line passing through $(3,2)$ having slope $\frac{3}{4}$ is given by
$y-2=\frac{3}{4}(x-3)$
$\Rightarrow 4y-8=3x-9$
$\Rightarrow 4y-3x+1=\mathrm{0.......}$(i)
Let $(h,k)$ be the point on the line which is 5 units away from the point $A$. Then,
$\sqrt{(h-3{)}^2+(k-2{)}^2}=\mathrm{5......}$(ii)
Also, we have
$4k-3h+1=0$ [from Eq. (i)]
$\Rightarrow k=\frac{3h-1}{4}......$(iii)
Putting the value of $k$ in Eq. (ii) we get
$\sqrt{(h-3{)}^2+{\left(\frac{3h-1}{4}-2\right)}^2}=5$
$\Rightarrow 16(h-3{)}^2+(3h-9{)}^2=400$
$\Rightarrow 16h^2+144-96h+9h^2+81-54h=400$
$\Rightarrow 25h^2-150h-175=0$
$\Rightarrow h^2-6h-7=0$
$\Rightarrow (h+1)(h-7)=0$
$\Rightarrow h=-1,h=7$
Putting these value of $h$ in Eq. (iii), we get
$k=\frac{3(-1)-1}{4}$
$\Rightarrow k=-1$
and $k=\frac{3(7)-1}{4}$
$\Rightarrow k=5$
Therefore, the coordinates of the required points are either $(-1,-1)$ or $(7,5)$.