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9  Text Solution Solution : Let the point be A(5,−6), B(−1,−4), and P(0,y) <br> Point P is on the y−axis, hence its x coordinate is 0. So, it is of the form P(0,y). <br> Now, we have to find the ratio. Let the ratio be k:1 Hence `m_1=k, m_2=1, x_1 =5, y_1 =−6, x_2=−1,y_2=−4, x=0 ` <br> Using sections formula `x= (m_1x_1+m_2x_2)/(m_1+m_2)` <br> `⇒0= (-k+5)/(k+1)` <br> `∴k=5` Again, `y= (m_1y_1+m_2y_2)/(m_1+m_2)` <br> `(−4k−6)/(k+1) = (−20−6)/6` <br> for `k=5` <br> `= -13/3` <br> Hence the coordiantes of the point is `P(0,-13/3)`
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Example 9 Find the ratio in which the y−axis divides the line segment joining the points (5, – 6) and (–1, – 4). Also find the point of intersection. Let the point be A(5, −6) & B(−1, −4) Let Point P the required required point Since Point P is on y−axis, hence its x coordinate is 0. So, it is of the form P(0, y) Now, we have to find ratio Let ratio be k : 1 Hence, m1 = k, m2 = 1 x1 = 5, y1 = −6 x2 = −1, y2 = −4 x = 0, y = y Using section formula x = (𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1+ 𝑚_2 ) 0 = (𝑘 ×−1 + 1 × 5)/(𝑘 + 1) 0 = (−𝑘 + 5)/(𝑘 + 1) 0(k + 1) = −k + 5 0 = −k + 5 k = 5 Hence, k = 5 Now, we need to find y also y = (𝑚_1 𝑦_2 + 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 ) = (𝑘 × −4 + 1 × −6)/(𝑘 + 1) = (5 × −4 + 1 × 1)/(5 + 1) = (−20 − 6)/6 = (−26)/6 = (−13)/3 Hence the coordinate of point is P(0, y) = P ("0, " (−𝟏𝟑)/𝟑) Answer Hint: In this question, we will use the concept of section formulae of coordinate geometry. This states that the coordinate of the point which divides the line segment joining the points $({x_1},{y_1})$and $({x_2},{y_2})$ internally in the ratio m:n is given by $\left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$. …….(i) Complete step-by-step answer: Here , we have given points of the line segment , (5,-6) and (-1,-4)Comparing this with $({x_1},{y_1})$and $({x_2},{y_2})$, we get ${x_1} = 5,{x_2} = - 1,{y_1} = - 6$ and ${y_2} = - 4$.Let the required ratio be k:1. Then by using the section formula,$\left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$According to the question, the point of division is on the y-axis, hence the x- coordinate of that point is 0.Then, the coordinates of the point of division are,$ \Rightarrow x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}} \\ \Rightarrow 0 = \dfrac{{k( - 1) + 1(5)}}{{k + 1}} \\ \Rightarrow 0 = 5 - k \\ \Rightarrow k = 5 \\ $Hence, the ratio in which the y-axis divides the line segment is 5:1.Now we have to find the coordinates of the point of division on y-axis, i.e. (0,y)$ \Rightarrow $$y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}$$ \Rightarrow y = \dfrac{{5( - 4) + 1( - 6)}}{{5 + 1}} \\ \Rightarrow y = \dfrac{{ - 20 - 6}}{6} = \dfrac{{ - 26}}{6} \\ \Rightarrow y = \dfrac{{ - 13}}{3} \\ $Hence, the coordinates of the point of division is $\left( {0,\dfrac{{ - 13}}{3}} \right)$.Note: In this type of question we have to remember the concept of the section formulae .First we have to find out the required values and then we will assume that the line segment will be divided in k:1. After that we will find out the coordinates of the point of division by putting those values in section formulae i.e. $\left( {x = \dfrac{{m{x_2} + n{x_1}}}{{m + n}},y = \dfrac{{m{y_2} + n{y_1}}}{{m + n}}} \right)$. After solving the formula step by step, we will get the required ratio of division and the coordinates of point of division.
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In what ratio does Y axis divides the line segment joining the points 5 and (- 1?
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