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  • The line  is rotated about its point of intersection with <img alt="3 x-y-10=0" src="https://ent

The line 2 x-y-7=0 is rotated about its point of intersection with 3 x-y-10=0 through an angle \frac{\pi}{4} in the anticlockwise direction. Find the point on the line in the new position which is at the same distance from the point of intersection of the given two lines as the origin.

Option: 1

\left(\frac{13}{3}, \frac{1}{2}\right)


Option: 2

\left(\frac{13}{2}, \frac{3}{2}\right)


Option: 3

\left(\frac{13}{6}, {3}\right)


Option: 4

\left(\frac{13}{6}, \frac{3}{2}\right)


Answers (1)

best_answer

Given lines are
\mathrm{\begin{aligned} & 2 x-y-7=0 \\ & 3 x-y-10=0 \end{aligned} }
Point of intersection of the lines is P(3,-1). Line (1) is rotated through an angle \mathrm{\frac{\pi}{4} } in anticlockwise direction about point \mathrm{\mathrm{P} }  and let it become line (3) with slope \mathrm{\mathrm{m}. }
\mathrm{\Rightarrow \quad \tan 45^{\circ}=\frac{m-2}{1+2 m} \quad \Rightarrow \quad 1+2 m=m-2 }
\mathrm{\begin{array}{ll} \Rightarrow & \mathrm{m}=-3 \\ \Rightarrow & \text { equation of line }(3) \text { is }(\mathrm{y}-1)=-3(\mathrm{x}-3) \\ \Rightarrow & 3 \mathrm{x}+\mathrm{y}=8 \end{array} }
Let \mathrm{Q(\alpha, \beta) } be a point on line (3) \mathrm{\Rightarrow \beta=(8-3 \alpha) } From the given condition of the problem

\mathrm{\begin{array}{ll} & \alpha^2+\beta^2=(\alpha-3)^2+(\beta+1)^2 \\ \Rightarrow & \alpha^2+(8-3 \alpha)^2=(\alpha-3)^2+9(\alpha-3)^2 \\ \Rightarrow & \alpha^2+9 \alpha^2-48 \alpha+64=10 \alpha^2-60 \alpha+90 \\ \Rightarrow & 12 \alpha=90-64=26 \\ \Rightarrow & \alpha=\frac{13}{6} \\ \Rightarrow & \beta=8-\frac{13}{2} \\ \Rightarrow & \beta=\frac{3}{2} \\ \therefore & Q=\left(\frac{13}{6}, \frac{3}{2}\right) . \end{array} }

Posted by

Suraj Bhandari

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