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 with $PA=PB$. In an isosceles triangle,the median from

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

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(A) Let $M$ be the mid-point of $AB$. Since $PA=PB$,the triangle $PAB$ is an isosceles triangle with $PA=PB$. In an isosceles triangle,the median from the vertex $P$ to the base $AB$ is also the altitude to the base $AB$. Thus,$PM \perp AB$. The slope of the line $AB$ (which is $2x-y+3=0$ or $y=2x+3$) is $m_1 = 2$. Since $PM \perp AB$,the slope of $PM$ is $m_2 = -\frac{1}{m_1} = -\frac{1}{2}$. The line $PM$ passes through $P(1,2)$ and has a slope of $-\frac{1}{2}$. The equation of line $PM$ is $y-2 = -\frac{1}{2}(x-1)$,which simplifies to $2y-4 = -x+1$,or $x+2y-5=0$. The mid-point $M$ is the intersection of the lines $2x-y+3=0$ and $x+2y-5=0$. From the first equation,$y=2x+3$. Substituting this into the second equation: $x+2(2x+3)-5=0 $ $\Rightarrow x+4x+6-5=0 $ $\Rightarrow 5x+1=0 $ $\Rightarrow x = -\frac{1}{5}$. Now,$y = 2(-\frac{1}{5})+3 = -\frac{2}{5}+3 = \frac{13}{5}$. 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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