Math, asked by Anonymous, 7 months ago

find the ratio in which the line segment joining the points A(3,-3) and B(-2,7) is divided by x axis. also find the coordinates of the points of division.​

Answers

Answered by Cynefin
88

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Required Answer:

✒ Provided:-

  • Endpoints of line segment A(3,-3) and B(-2,7)
  • It is divided by x axis.

✒ To FinD:-

  • The ratio of division
  • Coordinates of dividing points....?

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How to solve?

We could solve this question by apply formula and get solution process, if we know the section formula.

  • The coordinates of the point which divides a line segment joining the points (x1, y1) and (x2, y2) internally into the ratio m:n are given by,

 { \boxed{ \rm{x =  \frac{mx_2 + nx_1}{m + n}  \: and  \:  y =  \frac{my_2 + ny_1}{m + n} }}}

☁️ So, Let's solve this question....

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Solution:

We have,

  • Line segment joining points A(3,-3) & B(-2,7)

[ The point which will divide the segment lies in x axis, So let the point be (x,0) and let the ratio in which it divide is k:1 ]

⚘ Compare the ratio k:1 with m:n,

By using section formula,

{ \rm{p(x \: 0)=  \dfrac{k( - 2)+ 3}{k  + 1}   \:  \:  \dfrac{k( 7) +  - 3}{k + 1} }}

{ \rm{p(x \: 0)=  \dfrac{ - 2k+ 3}{k  + 1}   \:  \:  \dfrac{ 7k   - 3}{k + 1} }}

Comparing both sides,

{ \rm{x=  \dfrac{ - 2k+ 3}{k  + 1} ..........(1)  }}

{ \rm{ 0=   \dfrac{  7k  - 3}{k + 1}........(2) }}

From equation (2),

 \rm{  7k  - 3 = 0}

 \rm{k =  \dfrac{3}{7} }

Putting the value of k in equation (1),

{ \rm{x=  \dfrac{ - 2( \dfrac{3}{7}) + 3}{ \dfrac{3}{7}   + 1}  }}

 \rm{x =  \dfrac{  \dfrac{ - 6}{7} + 3 }{ \dfrac{10}{7} } }

 \rm{ x=  \dfrac{ \dfrac{15}{7} }{ \dfrac{10}{7} } }

 \rm{x =  \dfrac{3}{2} }

So, The required ratio is:

  • k : 1 = 3/7 : 1 = 3 : 7

And, the Required coordinates are:

  • P(x,0) = (3/2,0)

Hence, solved !!

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Answered by BrainlyEmpire
99

\huge\pink{\underline{\boxed{\mathbb{AnSwEr}}}}]

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\color{green} {} Heya!!

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\huge\red{\underline{\boxed{\mathbb{SoLuTIoN}}}}]

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✎Let the ratio be k : 1 .

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✎Then by the section formula, the coordinates of the point which divides AB in the ratio k : 1 are

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✎[ (-2k+3) / (k+1) , (7k - 3) / (k+1) ]

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✎The point lies on x-axis, and we know that on the x-axis the ordinate is 0.

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✎Therefore, (7k-3) / (k+1) = 0

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=> 7k-3 = 0

=> 7k = 3

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=> k = 3/7

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=> k : 1 = 3 : 7

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✎Putting the value of k = 3/7, we get point of intersection as

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✎{ [ -2(3/7) + 3] ÷ (3/7)+1 , 0 }

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=> { [(-6/7) + 3] ÷ (3/7) + 1 , 0 }

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=> [(-6+21)/7 ÷ (3+7)/7 , 0 ]

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=> [ 15/7 ÷ 10/7 , 0 ]

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=> [ 15/10 , 0 ]

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=> ( 3/2 , 0 ).

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\color{orange} {} Coordinates

\color{red} {} ( 3/2 , 0 ).

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