Question

Find the coordinates of point on the line joining the points p(3,-4) and q(-2,5) that is twice as far from p as from q.

Answer 1

Answer:

The coordinates of the point is $(-\frac{1}{3}, 2)$

Step-by-step explanation:

Given coordinates of P = (3, -4)

Coordinates of Q = (-2, 5)

We have to fine the coordinates of the point that is dividing the line joining P and Q in the ration 2 : 1

Let the coordinates of the point be (x, y)

If coordinates of P be $(x_1, y_1)$ and coordinates of Q be $(x_2, y_2)$ and the ratio in which P and Q is divided by a point (x, y) is $m_1 : m_2$ then

$x=\frac{m_1x_2+m_2x_1}{m_1+m_2}$

And, $y=\frac{m_1y_2+m_2y_1}{m_1+m_2}$

Thus,

$x=\frac{2\times (-2)+1\times 3}{1+2}$

$\implies x=\frac{2\times (-2)+1\times 3}{2+1}$

$\implies x=\frac{-4+3}{1+2}$

$\implies x=\frac{-1}{3}$

$y=\frac{2\times 5+1\times (-4)}{2+1}$

$\implies y=\frac{10-4}{3}$

$\implies y=\frac{6}{3}$

$\implies y=2$

Therefore, the coordinates of the point is $(-\frac{1}{3}, 2)$

Hope this helps.

Answer 2

Answer:

$(-\dfrac{1}{3},2)$

Step-by-step explanation:

The given points are p(3,-4) and q(-2,5).

It is given point on the line joining the points p(3,-4) and q(-2,5) is twice as far from p as from q.

It means the point divides the line segment pq in 2:1.

Section formula:

$(\dfrac{mx_2+nx_1}{m+n},\dfrac{my_2+ny_1}{m+n})$

Using section formula we get

$(\dfrac{(2)(-2)+(1)(3)}{2+1},\dfrac{(2)(5)+(1)(-4)}{2+1})$

$(\dfrac{-4+3}{3},\dfrac{10-4}{3})$

$(-\dfrac{1}{3},\dfrac{6}{3})$

$(-\dfrac{1}{3},2)$

Therefore, the coordinates of required point are $(-\dfrac{1}{3},2)$.