
By what ratio is the line segment joining (0, 2) and (6,8) divided by the line \[2x+2y=5\]?
Answer
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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.
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