
By what ratio is the line segment joining (0, 2) and (6,8) divided by the line \[2x+2y=5\]?
Answer
627k+ views
Hint: We will use the concept of internal divisions with section formula to solve this question. We will find the point of intersection between line segment AB and the given line equation \[2x+2y=5\] using the section formula \[P=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right)\] and then substitute x-coordinate and y-coordinate of the point P in the line equation.
Complete step-by-step answer:
Before proceeding with the question we should know about the concept of internal divisions with the section formula.
If point P(x,y) lies on line segment AB between points A and B and satisfies \[\text{AP:PB=m:n}\], then we say that P divides AB internally in the ratio m:n. The point of division has the coordinates
\[P=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right).........(1)\]
Let the points of the line segment AB are A(0,2) and B(6,8).
The given line in the question is \[2x+2y=5\].
Let the line \[2x+2y=5\] divide line segment AB in the ratio \[k:1\] at point P.
From equation (1) we get the coordinates of point P \[=\left( \dfrac{6k+0}{k+1},\,\dfrac{8k+2}{k+1} \right)\]
From the question we can conclude that point P lies on the line \[2x+2y=5.......(2)\]
So now substituting \[x=\dfrac{6k}{k+1}\] and \[y=\dfrac{8k+2}{k+1}\] in equation (2) we get,
\[\,\Rightarrow 2\left( \dfrac{6k}{k+1} \right)+2\left( \dfrac{8k+2}{k+1} \right)=5.......(3)\]
Rearranging and simplifying equation (3) we get,
\[\,\Rightarrow \dfrac{12k}{k+1}+\dfrac{16k+4}{k+1}=5.......(4)\]
Taking L.C.M in equation (4) we get,
\[\,\Rightarrow \dfrac{12k+16k+4}{k+1}=5.......(5)\]
Now cross multiplying and shifting similar terms to one side and the numbers to another side and then solving for k we get,
\[\begin{align}
& \,\Rightarrow 28k+4=5k+5 \\
& \,\Rightarrow 28k-5k=5-4 \\
& \,\Rightarrow 23k=1 \\
& \,\Rightarrow k=\dfrac{1}{23} \\
\end{align}\]
So the ratio will be $\dfrac{1}{23}:1$ which can be written as 1:23.
Hence the ratio in which the line segment AB gets divided by \[2x+2y=5\] is \[1:23\] internally.
Note: We will have to remember the section formula and the important thing here is to assume the ratio \[k:1\] because it will reduce one variable. We can substitute y coordinate in place of x coordinate in the section formula by mistake so we need to be careful while doing the substitution of coordinates in the section formula.
Complete step-by-step answer:
Before proceeding with the question we should know about the concept of internal divisions with the section formula.
If point P(x,y) lies on line segment AB between points A and B and satisfies \[\text{AP:PB=m:n}\], then we say that P divides AB internally in the ratio m:n. The point of division has the coordinates
\[P=\left( \dfrac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\,\dfrac{m{{y}_{2}}+n{{y}_{1}}}{m+n} \right).........(1)\]
Let the points of the line segment AB are A(0,2) and B(6,8).
The given line in the question is \[2x+2y=5\].
Let the line \[2x+2y=5\] divide line segment AB in the ratio \[k:1\] at point P.
From equation (1) we get the coordinates of point P \[=\left( \dfrac{6k+0}{k+1},\,\dfrac{8k+2}{k+1} \right)\]
From the question we can conclude that point P lies on the line \[2x+2y=5.......(2)\]
So now substituting \[x=\dfrac{6k}{k+1}\] and \[y=\dfrac{8k+2}{k+1}\] in equation (2) we get,
\[\,\Rightarrow 2\left( \dfrac{6k}{k+1} \right)+2\left( \dfrac{8k+2}{k+1} \right)=5.......(3)\]
Rearranging and simplifying equation (3) we get,
\[\,\Rightarrow \dfrac{12k}{k+1}+\dfrac{16k+4}{k+1}=5.......(4)\]
Taking L.C.M in equation (4) we get,
\[\,\Rightarrow \dfrac{12k+16k+4}{k+1}=5.......(5)\]
Now cross multiplying and shifting similar terms to one side and the numbers to another side and then solving for k we get,
\[\begin{align}
& \,\Rightarrow 28k+4=5k+5 \\
& \,\Rightarrow 28k-5k=5-4 \\
& \,\Rightarrow 23k=1 \\
& \,\Rightarrow k=\dfrac{1}{23} \\
\end{align}\]
So the ratio will be $\dfrac{1}{23}:1$ which can be written as 1:23.
Hence the ratio in which the line segment AB gets divided by \[2x+2y=5\] is \[1:23\] internally.
Note: We will have to remember the section formula and the important thing here is to assume the ratio \[k:1\] because it will reduce one variable. We can substitute y coordinate in place of x coordinate in the section formula by mistake so we need to be careful while doing the substitution of coordinates in the section formula.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

The largest wind power cluster is located in the state class 11 social science CBSE

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

Which among the following are examples of coming together class 11 social science CBSE

