Q.15. The co-ordinates of the points on the line 2x-y = 5 which are the distance of 1 unit from the line 3x+4y=5 are
A. \left ( \frac{30}{11},\frac{-5}{11} \right ),\left ( \frac{20}{11},\frac{15}{11} \right )
B. \left ( -\frac{30}{11},\frac{-5}{11} \right ),\left ( -\frac{20}{11},\frac{15}{11} \right )
C. \left ( \frac{30}{11},\frac{5}{11} \right ),\left ( \frac{20}{11},\frac{-15}{11} \right )
D. \left ( \frac{-30}{11},\frac{5}{11} \right ),\left ( \frac{-20}{11},\frac{-15}{11} \right )
Answer :- C. \left ( \frac{30}{11},\frac{5}{11} \right ),\left ( \frac{20}{11},\frac{-15}{11} \right )
Explanation :-
\text{Let } (x_{1},y_{1}) \text{ be the required point} \therefore \quad 2x_{1}-y_{1}=5 \quad ......(\text{i}) \text{Also, } (x_1,y_1) \text{ is at the distance of 1 unit from the line } 3x+4y=5 \therefore \quad 1 = \left| \frac{3x_{1}+4y_{1}-5}{\sqrt{9+16}} \right| \therefore \quad \pm5 = 3x_1+4y_1-53x1 + 4y1 -5 =5 or 3x1 4y1 -5 = -5
3x1+4y1 = 10 —–(ii)
or
3x1+4y1 = 0 —–(iii)
Solving eq. (i) and (ii) , we get
\; x_{1}=\frac{30}{11} \; and\; y_{1}=\frac{5}{11}Solving equations (i) and (iii), we get
\; x_{1}=\frac{20}{11} \; and\; y_{1}=\frac{-15}{11}Therefore, \left ( \frac{30}{11},\frac{5}{11} \right ) and \left ( \frac{20}{11},\frac{-15}{11} \right ) are the required points.