Points Q, R, and S lie on the line segment jo | vedclass

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(B) Let the points be $P, Q, R, S, T$ on a line such that $PQ = QR = RS = ST = k$. The coordinates of $P$ are $(a, x)$ and $T$ are $(b, y)$. The total

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$QR$

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(B) Let the points be $P, Q, R, S, T$ on a line such that $PQ = QR = RS = ST = k$. The coordinates of $P$ are $(a, x)$ and $T$ are $(b, y)$. The total length $PT = 4k$. The point $L$ has coordinates $\left( \frac{5a + 3b}{8}, \frac{5x + 3y}{8} ight)$. Using the section formula,a point dividing $PT$ in the ratio $m:n$ has coordinates $\left( \frac{mb + na}{m+n}, \frac{my + nx}{m+n} ight)$. Comparing this with $L$,we have $\frac{n}{m+n} = \frac{5}{8}$ and $\frac{m}{m+n} = \frac{3}{8}$,which gives the ratio $m:n = 3:5$. Thus,$L$ divides $PT$ in the ratio $3:5$. Since $PQ = k, QR = k, RS = k, ST = k$,the distance $PL = \frac{3}{8} 	imes 4k = 1.5k$. This means $L$ lies between $Q$ and $R$ such that $QL = 0.5k$ and $LR = 0.5k$. Therefore,$L$ is the midpoint of $QR$.

Points $Q, R,$ and $S$ lie on the line segment joining $P(a, x)$ and $T(b, y)$ such that $PQ = QR = RS = ST$. If $L\left( \frac{5a + 3b}{8}, \frac{5x + 3y}{8} \right)$ is a point on the segment,then $L$ is the midpoint of which line segment?

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