Question

The distance of the point of intersection of the lines 2x – 3y + 5 = 0 are 3x + 4y = 0 from the line 5x – 2y = 0, is

(a) $\frac{130}{17\sqrt{29}}$

(b) $\frac{13}{7\sqrt{29}}$

(a) $\frac{130}{7}$

(d) none of these

Answer 1

Point of intersection of 2x − 3y + 5 = 0 and 3x + 4y = 0 is

$$\begin{gathered} \mathrm{i}.\mathrm{e}.\begin{array}{c}3\times \\ 2\times \end{array}\left[\begin{array}{l}2x-3y+5=0\\ 3x+4y=0\end{array}\right. \\ \frac{\begin{array}{l}\mathrm{i}.\mathrm{e}6x-9y+15=0\\ 6x+8y=0\\ --\end{array}}{\mathrm{i}.\mathrm{e}.-17y+15=0} \\ \mathrm{if}y=\frac{15}{17} \\ \therefore 3x=-4\left(\frac{15}{17}\right) \\ \mathrm{i}.\mathrm{e}.x=\frac{-4\times 15}{17\times 3}=\frac{-20}{17} \\ \therefore \mathrm{Distance}\mathrm{of}\left(\frac{-20}{17},\frac{15}{17}\right)\mathrm{from}5x-2y=0\mathrm{is}\frac{\left|5\left({\displaystyle \frac{-20}{17}}\right)-2\left({\displaystyle \frac{15}{17}}\right)\right|}{\sqrt{25+4}} \\ \mathrm{i}.\mathrm{e}.\frac{1}{\sqrt{29}}\left[\frac{\left|-100-30\right|}{17}\right]=\frac{130}{17\sqrt{29}} \end{gathered}$$

Hence, the correct answer is option A.