Question

Calculate the ratio in which the line joining $A(-4,2)$ and $B(3,6)$ is divided by a point $P(x,3)$. Also find
(i) $x$.
(ii) length of $AP$.

Answer 1

Step 1: Let $P(x,3)$ divides the line joining in the ratio$k:1$.

Using section formula;

$Coordinatesofmid-point=\left(\frac{m_1{x}_2+m_2{x}_1}{m_2+m_1},\frac{m_1{y}_2+m_2{y}_1}{m_2+m_1}\right)$

Now, compare this with the given question we see that
$a=x,b=3,x_1=-4,y_1=2,x_2=3,y_2=6,m_1=k,m_2=1$

Step 2: Solve for $x$ and $y$:
$x=\frac{3\times k-4\times 1}{k+1} -\left(1\right)$

Solve for $y$:
$3=\frac{6\times k+2\times 1}{k+1} -\left(2\right)$

$3(k+1)=6k+2$

$\Rightarrow$ $3k=1$

$\Rightarrow$ $k=\frac{1}{3}$

Substituting the value of $k$ in eq-(1) we have
$x=\frac{3\times \frac{1}{3}-4\times 1}{\frac{1}{3}+1}$

$\Rightarrow \frac{-9}{4}=x$

Step 3: Now let us find distance using distance formula

$D=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$
$AP=\sqrt{{(-\frac{9}{4}+4)}^2+{(3-2)}^2}$

$AP=\sqrt{{\left(\frac{7}{4}\right)}^2+{\left(1\right)}^2}=\sqrt{\frac{49}{16}+1}=\sqrt{\frac{65}{16}}$

Thus,

(i) The coordinates of $P$ are ($-\frac{9}{4},3$)
(ii) The distance $AP$ is $\frac{\sqrt{65}}{4}$ units.