Suppose A and B are two points on the line 2x | vedclass

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(A) Let $M$ be the mid-point of $AB$. Since $PA = PB$,the triangle $PAB$ is an isosceles triangle,and the line segment $PM$ is the perpendicular bisec

Vedclass · Accepted Answer

$\left(\frac{-1}{5}, \frac{13}{5}\right)$

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(A) Let $M$ be the mid-point of $AB$. Since $PA = PB$,the triangle $PAB$ is an isosceles triangle,and the line segment $PM$ is the perpendicular bisector of $AB$.Given the line $L: 2x - y + 3 = 0$,its slope is $m_L = 2$.Since $PM \perp AB$,the slope of $PM$ is $m_{PM} = -\frac{1}{m_L} = -\frac{1}{2}$.The line $PM$ passes through $P(1, 2)$ and has a slope of $-\frac{1}{2}$. Its equation is:$y - 2 = -\frac{1}{2}(x - 1)$$2y - 4 = -x + 1$$x + 2y = 5$  (Equation $ii$)The mid-point $M$ is the intersection of the line $2x - y = -3$ (Equation $i$) and $x + 2y = 5$ (Equation $ii$).From $(i)$,$y = 2x + 3$. Substituting into (ii):$x + 2(2x + 3) = 5$$x + 4x + 6 = 5$$5x = -1 \implies x = -\frac{1}{5}$$y = 2(-\frac{1}{5}) + 3 = -\frac{2}{5} + \frac{15}{5} = \frac{13}{5}$Thus,the mid-point $M$ is $\left(-\frac{1}{5}, \frac{13}{5}\right)$.

Suppose $A$ and $B$ are two points on the line $2x - y + 3 = 0$ and $P(1, 2)$ is a point such that $PA = PB$. Then,the mid-point of $AB$ is

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