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Question

Find the distance of the line $$4x + 7y + 5$$ $$\displaystyle = $$ $$0$$ from the point $$(1, 2)$$ along the line $$2x - y \displaystyle = $$ $$0$$.


Solution

The given lines are
$$\displaystyle 2x-y=0 $$...(1)
$$\displaystyle 4x+7y+5=0 $$....(2)
$$A(1,2)$$ is point on line (1).
Let $$B$$ be the point of intersection of lines (1) and (2).
On solving equation (1) and (2), we obtain$$\displaystyle x= \frac{-5}{18}\text{and}  y= \frac{-5}{9}$$
$$\displaystyle \therefore $$coordinates of point $$B$$ are$$\displaystyle \left \{ \frac{-5}{18},\frac{-5}{9} \right \}$$
By using distance formula, the distance between point $$A$$ and $$B$$ can be obtained as$$\displaystyle AB= \sqrt{\left \{ 1+\frac{5}{18} \right \}^{2}+\left \{ 2+\frac{5}{9} \right \}^{2}}units$$
$$\displaystyle = \sqrt{\left \{ \frac{23}{18} \right \}^{2}+\left \{ \frac{23}{9} \right \}^{2}}units$$
$$\displaystyle = \sqrt{\left \{ \frac{23}{2\cdot 9} \right \}^{2}+\left \{ \frac{23}{9} \right \}^{2}}units$$
$$\displaystyle = \sqrt{\left \{ \frac{23}{9} \right \}^{2}\left \{ \frac{1}{2} \right \}^{2}+\left \{ \frac{23}{9} \right \}^{2}}units
$$
$$\displaystyle = \sqrt{\left ( \frac{23}{9} \right )^{2}\left ( \frac{1}{4}+1 \right )}units$$
$$\displaystyle = \frac{23}{9}\sqrt{\frac{5}{4}}units$$
$$\displaystyle = \frac{23}{9}\cdot\frac{\sqrt{5}}{2}units$$
$$\displaystyle = \frac{23\sqrt{5}}{18}units$$
Thus, the required distance is$$\displaystyle = \frac{23\sqrt{5}}{18}units$$

Mathematics
NCERT
Standard XI

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