Question

A(0,3),B(-1,-2) and C(4,2) are the vertices of a triangle abc. D is a point on the side BC such that BD/DC =1 /2.P is a point on AD such that AP =2root5/3 units. Find the coordinates of P

Answer 1

Answer:  The answer is $\left(\dfrac{4}{15},\dfrac{23}{15}\right)$.

Step-by-step explanation:  Given that the vertices of a ΔABC are A(0, 3), B(-1, -2) and C(4, 2). 'D' is a point on BC such that BD : DC = 1 : 2.

Also, 'P' is a point on AD, where $AP=\dfrac{2\sqrt5}{3}.$ We are to find the co-ordinates of the point 'P'.

The figure is attached.

Division formula: We know that if a point Q divides a line segment joining the points (a, b) and (c, d) in the ratio m : n internally, the the co-ordinates of Q are given by

$\left(\dfrac{mc+na}{m+n},\dfrac{md+nb}{m+n}\right).$

Since point D divides BC in the ratio 1 : 2, so its co-ordinates will be

$D=\left(\dfrac{1\times 4+2\times (-1)}{1+2},\dfrac{1\times 2+2\times (-2)}{1+2}\right)\\\\\\\Rightarrow D=\left(\dfrac{2}{3},-\dfrac{2}{3}\right).$

Therefore, the length of AD is given by

$AD=\sqrt{\left(0-\dfrac{2}{3}\right)^2+\left(3+\dfrac{2}{3}\right)^2}=\sqrt{\left(\dfrac{4}{9}\right)+\left(\dfrac{121}{9}\right)}=\sqrt{\dfrac{125}{9}}=\dfrac{5\sqrt5}{3}.$

Hence the length of PD will be

$PD=AD-AP=\dfrac{5\sqrt5}{3}-\dfrac{2\sqrt5}{3}=\sqrt5.$

Therefore, the ratio in which P divides AD is given by

$\dfrac{m_1}{n_1}=\dfrac{\frac{2\sqrt5}{3}}{\sqrt5}=\dfrac{2}{3}\\\\\Rightarrow m_1:n_1=2:3.$

Thus, the co-ordinates of point P are

$P\\\\=\left(\dfrac{2\times\frac{2}{3}+3\times 0}{2+3},\dfrac{2\times(-\frac{2}{3})+3\times 3}{2+3}\right)\\\\\\=\left(\dfrac{\frac{4}{3}}{5},\dfrac{\frac{23}{3}}{5}\right)\\\\\\=\left(\dfrac{4}{15},\dfrac{23}{15}\right).$

The answer is $\left(\dfrac{4}{15},\dfrac{23}{15}\right)$.