2
$\begingroup$

The lines $y= 4x + 2$ and $x+2y=6$ intersect at point $P$.

i) Find the coordinates of point $P$.

$$y= 4x + 2\ldots(1)$$ $$x+2y=6 \ldots(2)$$

Sub ($1$) into ($2$)

$$x+2(4x + 2) =6$$

$$9x=2$$

$$x=2/9$$

sub $x = 2/9$ into ($1$) to find $y$

$$y=26/9$$

Therefore the coordinates of $P$ is $(\frac{2}{9}, \frac{26}{9})$.

ii) Find the ratio in which $P$ divides the interval $(1,\frac{33}{9})$ and $(\frac{1}{3},3)$

Using the interval division formula, and letting the ratio be $k:l$, I got two equations, $\frac{k (1/3) + l(1) }{k+l}$ and $\frac{k (3) + l(33/9) }{k+l}$.

However when I simplified both, I ended up getting $7l+k=0$ on both sides, which left me stuck as I was not sure where I went wrong. Should I try with the ratio $-k:l$?

$\endgroup$
1
$\begingroup$

HINT:

The distance between two points $(x_1,y_1), \; (x_2,y_2)$ is:

$\sqrt{(x_1-x_2)^2+ (y_1-y_2)^2}$.

In your case, let $A(1,\frac{33}{9}),\; B(\frac{1}{3},3)$.

Then $PA=\frac{7}{9}\sqrt 2$ and $PB=\frac{1}{9}\sqrt 2$.

Thus, $\frac{PA}{PB}=7$.

$\endgroup$
0
$\begingroup$

Notice,let $m:n$ be the ration in which the given point $P\left(\frac{2}{9}, \frac{26}{9}\right)$ divides the line joining the points $\left(1, \frac{33}{9}\right)$ & $\left(\frac{1}{3}, 3\right)$ respectively

then the coordinates of the point P are calculated using division formula as follows $$P\equiv \color{red}{\left(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n}\right)}$$ Now, substituting the corresponding values we get $$\left(\frac{2}{9}, \frac{26}{9}\right)\equiv \left(\frac{\frac{1}{3}\cdot m+1\cdot n}{m+n}, \frac{3\cdot m+n\cdot \frac{33}{9}}{m+n}\right)$$

Now, comparing the corresponding $x$ coordinates, we get $$\frac{\frac{1}{3}\cdot m+1\cdot n}{m+n}=\frac{2}{9}\iff m+7n=0$$ $$\frac{m}{n}=-\frac{7}{1}$$ Similarly, comparing the y-coordinate, we get
$$\frac{3\cdot m+n\cdot \frac{33}{9}}{m+n}=\frac{26}{9}\iff m+7n=0$$ $$\frac{m}{n}=-\frac{7}{1}$$ Negative sign indicates that the point $P$ divides the interval $\left(1, \frac{33}{9}\right)$ & $\left(\frac{1}{3}, 3\right)$ externally in a ratio $7:1$

Thus, we conclude that $$\bbox[5pt, border:2.5pt solid #FF0000]{\color{green}{\text{Point P divides the given interval externally in a ratio}\ \ \color{blue}{7:1}}}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.