Question

A line y = mx through origin cuts the parallel lines x + y = 3 and x + y = 5 at A and B respectively. The distance between the two points of intersection is d. If AB = 2 then the quadratic equation formed is
a) (4 - d²)m² + 2md² + 4 - d² = 0
b) (4 - d²)m² + 2md² + d² - 4 = 0
c) (4 - d²)m² - 2md² + 4 - d² = 0
d) None of these

Answer 1

$$\begin{gathered} \mathrm{Hi}, \\ \mathrm{Solving}\mathrm{y}=\mathrm{mx}\mathrm{with}\mathrm{x}+\mathrm{y}=3 \\ \mathrm{x}+\mathrm{mx}=3 \\ \mathrm{x}=\frac{3}{\mathrm{m}+1} \\ \mathrm{so}\mathrm{y}=\frac{3\mathrm{m}}{\mathrm{m}+1} \\ \mathrm{so}\mathrm{A}=\left(\frac{3}{\mathrm{m}+1},\frac{3\mathrm{m}}{\mathrm{m}+1}\right) \\ \mathrm{and}\mathrm{Solving}\mathrm{y}=\mathrm{mx}\mathrm{with}\mathrm{x}+\mathrm{y}=5 \\ \mathrm{x}+\mathrm{mx}=5 \\ \mathrm{x}=\frac{5}{\mathrm{m}+1} \\ \mathrm{so}\mathrm{y}=\frac{5\mathrm{m}}{\mathrm{m}+1} \\ \mathrm{so}\mathrm{B}=\left(\frac{5}{\mathrm{m}+1},\frac{5\mathrm{m}}{\mathrm{m}+1}\right) \\ \mathrm{now}\mathrm{d}=\mathrm{AB}=\sqrt{{\left(\frac{5}{\mathrm{m}+1}-\frac{3}{\mathrm{m}+1}\right)}^2+{\left(\frac{5\mathrm{m}}{\mathrm{m}+1}-\frac{3\mathrm{m}}{\mathrm{m}+1}\right)}^2} \\ =\sqrt{\frac{4}{{\left(\mathrm{m}+1\right)}^2}+\frac{4{\mathrm{m}}^2}{{\left(\mathrm{m}+1\right)}^2}}=\sqrt{\frac{4+4{\mathrm{m}}^2}{{\left(\mathrm{m}+1\right)}^2}} \\ {\mathrm{d}}^2=\frac{4+4{\mathrm{m}}^2}{{\left(\mathrm{m}+1\right)}^2} \\ {\mathrm{d}}^2\left(1+{\mathrm{m}}^2+2\mathrm{m}\right)=4+4{\mathrm{m}}^2 \\ \left(4-{\mathrm{d}}^2\right){\mathrm{m}}^2-2{\mathrm{md}}^2+4-{\mathrm{d}}^2=0 \\ \mathrm{so}\mathrm{here}\mathrm{AB}=2\mathrm{has}\mathrm{no}\mathrm{role} \end{gathered}$$