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$$\displaystyle = \frac{23\sqrt{5}}{18}units$$MathematicsNCERTStandard XI

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