Question

7. A point P on the Y axis divides the line segment joining the points (4,5) and (-3,3) in a
certain ratio. Find the coordinates of the point P

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$Let\: the \: ratio \: be \: k:1 .$

$Then \: by\: the \; section \: formula ,\:the$

$coordinates \: of \:the \: point \: which$

$divides\: joining \: of \: the \: points \: (4,5)$

$and \: (-3,3) \: in \: the \: ratio \: k:1 \: are$

$Here, x_{1} = 4, y_{1} = 5 , x_{2} = -3, y_{2} = 3$

$\Big( \frac{kx_{2} + 1\times x_{1}}{k+1}, \frac{ky_{2} + 1\times y_{1}}{k+1}\Big)$

$= \Big( \frac{k\times (-3) + 4 }{k+1} , \frac{ k \times 3 + 5 }{k+1} \Big) \: --(1)$

$This \:point \: lies \:on \: the \: y - axis , and$

$we \: know \:that \: on \: the \: y-axis \: the$

$abscissa \: is \: '0'$

$\therefore \frac{-3k+1}{k+1} = 0$

$\implies -3k + 1 = 0$

$\implies -3k = -1$

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

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

$So,the \: ratio \: is \: k : 1 = 1 : 3$

/* Putting the value of k in equation (1) ,we get Coordinates of point P */

$P = \Big( \frac{\frac{1}{3} \times (-3) + 4 }{\frac{1}{3} +1} , \frac{ \frac{1}{3} \times 3 + 5 }{\frac{1}{3}+1} \Big)$

$= \Big( \frac{ -3+12 }{1+3} , \frac{3+15}{1+3}\Big)$

$= \Big( \frac{9}{4} , \frac{18}{4}\Big)$

$= \Big( \frac{9}{4}, \frac{9}{2} \Big)$

Therefore.,

$\red{Coordinates \:of \:point\: P} \green { = \Big( \frac{9}{4}, \frac{9}{2} \Big)}$

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