Given two points Q(3,4) and R(1,2). What is t | vedclass

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(C) The condition $PQ+PR=QR$ implies that the point $P$ must lie on the line segment $QR$. First,we find the equation of the line passing through $Q(3

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$(2,3)$

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(C) The condition $PQ+PR=QR$ implies that the point $P$ must lie on the line segment $QR$. First,we find the equation of the line passing through $Q(3,4)$ and $R(1,2)$. The slope $m = \frac{2-4}{1-3} = \frac{-2}{-2} = 1$. The equation of the line $QR$ is $y - 2 = 1(x - 1)$,which simplifies to $x - y + 1 = 0$. Since $P$ lies on both the line $2x - y - 1 = 0$ and the line $x - y + 1 = 0$,we solve the system of equations: $2x - y = 1$ $x - y = -1$ Subtracting the second equation from the first gives $(2x - x) - (y - y) = 1 - (-1)$,so $x = 2$. Substituting $x = 2$ into $x - y = -1$,we get $2 - y = -1$,which implies $y = 3$. Thus,the point $P$ is $(2,3)$.

Given two points $Q(3,4)$ and $R(1,2)$. What is the point $P(x, y)$ on the line $2x-y-1=0$ for which $PQ+PR=QR$ holds?

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