Question

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:

Let the ratio in which line segment is divided be "k:1"

Then by the section formula,the coordinates of the point P are [(-3k+4)/(k+1) , (3k+5)/(k+1)]

This point lies on the y-axis and on y-axis,the abscissa is 0.

Therefore, (-3k+4)/(k+1) = 0

Solving this equation, we get k=4/3

That is,the ratio is 4:3.Putting the value of k=4,we get the point of intersection as (0,17/5)

So coordinates of pt. P are (0,17/5)

Answer 2

Answer:

here is your answer

Step-by-step explanation:

Lettheratiobek:1.

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

coordinates \: of \:the \: point \: whichcoordinatesofthepointwhich

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

and \: (-3,3) \: in \: the \: ratio \: k:1 \: areand(−3,3)intheratiok:1are

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

k+1

kx

2

+1×x

1

,

k+1

ky

2

+1×y

1

)

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

k+1

k×(−3)+4

,

k+1

k×3+5

)−−(1)

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

we \: know \:that \: on \: the \: y-axis \: theweknowthatonthey−axisthe

abscissa \: is \: '0'abscissais

′

0

′

\therefore \frac{-3k+1}{k+1} = 0∴

k+1

−3k+1

=0

\implies -3k + 1 = 0⟹−3k+1=0

\implies -3k = -1⟹−3k=−1

\implies k = \frac{-1}{-3}⟹k=

−3

−1

\implies k = \frac{1}{3}⟹k=

3

1

So,the \: ratio \: is \: k : 1 = 1 : 3So,theratioisk: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)P=(

3

1

+1

3

1

×(−3)+4

,

3

1

+1

3

1

×3+5

)

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

1+3

−3+12

,

1+3

3+15

)

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

4

9

,

4

18

)

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

4

9

,

2

9

)