
Find the ratio in which the point \[(11,15)\]divides the line segment joining the points \[(15,5)\]and \[(9,20)\]
Answer
522.3k+ views
Hint: In this question section formula will be used which tell us the coordinates of point which divides a given line segment into two parts in ratio \[m:n\]
\[x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}\]
\[y = \dfrac{{x{y_2} + n{y_1}}}{{m + n}}\]
Where \[({x_1}{y_1})\] and \[({x_2},{y_2})\]6- coordinates of points given.
Complete step by step answer:
Let ratio is \[k:1\]
Let point given are \[A(15,5)\] and \[B(9,20)\]
\[{x_1} = 15,{y_1} = 5\] and \[{x_2} = 9,{y_2} = 20\] and \[m:n\] is \[k:1\]
\[x = \dfrac{{k \times 9 + 1 \times 15}}{{K + 1}},\,y = \dfrac{{k \times 20 + 1 \times 5}}{{K + 1}}\]
\[x = \dfrac{{9k + 15}}{{K + 1}},\,y = \dfrac{{20k + 5}}{{K + 1}}\]
Points \[(11,15)\] divides line joining points \[(15,5)\]and \[(9,20)\]. Here \[x = 11,y = 15\]
So, \[\dfrac{{9k + 15}}{{k + 1}} = 11\]
\[ \Rightarrow 9k + 15 = 11k + 11\]
\[ \Rightarrow 9k - 11k = 11 - 15\]
\[ \Rightarrow - 2k = - 4\]
\[ \Rightarrow k = 2\]
Ration is \[2:1\]
Note: The intercept theorem is about the ratio of line segments. We have two lines intersecting in point S. Let two parallel lines intersect them in points A, B, C, and D. The points make up various lines. segments such as (the line from S to A), (the line from A to C), and so on. The theorem tells us about the ratios of the lengths of those line segments. That is useful if we know some of them but not all, then we can use the intercept theorem and solve it for the line segment.
here in this question we can also points \[y = 15\]equals to \[\dfrac{{20k + 15}}{{k + 1}}\] and get \[k = 2\]
\[x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}}\]
\[y = \dfrac{{x{y_2} + n{y_1}}}{{m + n}}\]
Where \[({x_1}{y_1})\] and \[({x_2},{y_2})\]6- coordinates of points given.
Complete step by step answer:
Let ratio is \[k:1\]
Let point given are \[A(15,5)\] and \[B(9,20)\]
\[{x_1} = 15,{y_1} = 5\] and \[{x_2} = 9,{y_2} = 20\] and \[m:n\] is \[k:1\]
\[x = \dfrac{{k \times 9 + 1 \times 15}}{{K + 1}},\,y = \dfrac{{k \times 20 + 1 \times 5}}{{K + 1}}\]
\[x = \dfrac{{9k + 15}}{{K + 1}},\,y = \dfrac{{20k + 5}}{{K + 1}}\]
Points \[(11,15)\] divides line joining points \[(15,5)\]and \[(9,20)\]. Here \[x = 11,y = 15\]
So, \[\dfrac{{9k + 15}}{{k + 1}} = 11\]
\[ \Rightarrow 9k + 15 = 11k + 11\]
\[ \Rightarrow 9k - 11k = 11 - 15\]
\[ \Rightarrow - 2k = - 4\]
\[ \Rightarrow k = 2\]
Ration is \[2:1\]
Note: The intercept theorem is about the ratio of line segments. We have two lines intersecting in point S. Let two parallel lines intersect them in points A, B, C, and D. The points make up various lines. segments such as (the line from S to A), (the line from A to C), and so on. The theorem tells us about the ratios of the lengths of those line segments. That is useful if we know some of them but not all, then we can use the intercept theorem and solve it for the line segment.
here in this question we can also points \[y = 15\]equals to \[\dfrac{{20k + 15}}{{k + 1}}\] and get \[k = 2\]
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