Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find the equation of the line.

Class 11 | Exercise 10.2 | Maths | NCERT | Straight Lines

Allow us to consider, AB be the line portion to such an extent that r (h, k) separates it in the proportion \[\mathbf{1}:\text{ }\mathbf{2}.\]

So the directions of An and B be (0, y) and (x, 0) separately.

We realize that the directions of a point isolating the line fragment joins the focuses (x1, y1) and (x2, y2) inside in the proportion m: n is

\[\mathbf{h}\text{ }=\text{ }\mathbf{2x}/\mathbf{3}\text{ }\mathbf{and}\text{ }\mathbf{k}\text{ }=\text{ }\mathbf{y}/\mathbf{3}\]

\[\mathbf{x}\text{ }=\text{ }\mathbf{3h}/\mathbf{2}\text{ }\mathbf{and}\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{3k}\]

\[\therefore \mathbf{A}\text{ }=\text{ }\left( \mathbf{0},\text{ }\mathbf{3k} \right)\text{ }\mathbf{and}\text{ }\mathbf{B}\text{ }=\text{ }\left( \mathbf{3h}/\mathbf{2},\text{ }\mathbf{0} \right)\]

We realize that the condition of the line going through the focuses \[\left( \mathbf{x1},\text{ }\mathbf{y1} \right)\text{ }\mathbf{and}\text{ }\left( \mathbf{x2},\text{ }\mathbf{y2} \right)\] is given by

\[\mathbf{3h}\left( \mathbf{y}\text{ }\text{ }\mathbf{3k} \right)\text{ }=\text{ }-\text{ }\mathbf{6kx}\]

\[\mathbf{3hy}\text{ }\text{ }\mathbf{9hk}\text{ }=\text{ }-\text{ }\mathbf{6kx}\]

\[\mathbf{6kx}\text{ }+\text{ }\mathbf{3hy}\text{ }=\text{ }\mathbf{9hk}\]

Allow us to partition both the sides by 9hk, we get,

\[\mathbf{2x}/\mathbf{3h}\text{ }+\text{ }\mathbf{y}/\mathbf{3k}\text{ }=\text{ }\mathbf{1}\]

∴ The equation is \[\mathbf{2x}/\mathbf{3h}\text{ }+\text{ }\mathbf{y}/\mathbf{3k}\text{ }=\text{ }\mathbf{1}\]

i

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Let $\begin{array}{l} f(x) = \left\{ \begin{array}{l} \frac{|x-3|}{{(x-3)’}},x \ne 3\\ 0,x = 3 \end{array} \right.\\ \end{array}$ . Show that $\begin{array}{l} \mathop {\lim }\limits_{x \to 3} f(x) \end{array}$ does not exist.

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