In square XYZW,overline{XY} parallel overline | vedclass

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(B) Given that $\overline{XY} \parallel \overline{ZW}$ in $\square XYZW$. Since $\overline{XZ}$ and $\overline{YW}$ are transversals intersecting at $

Vedclass · Accepted Answer

$2$

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(B) Given that $\overline{XY} \parallel \overline{ZW}$ in $\square XYZW$. Since $\overline{XZ}$ and $\overline{YW}$ are transversals intersecting at $P$,$	riangle PXY \sim 	riangle PZW$ by $AA$ similarity criterion (alternate interior angles are equal). Therefore,the ratios of their corresponding sides are equal: $\frac{PX}{PZ} = \frac{PY}{PW}$. Substituting the given values: $\frac{3x - 1}{5x - 3} = \frac{2x + 1}{6x - 5}$. Cross-multiplying: $(3x - 1)(6x - 5) = (2x + 1)(5x - 3)$. Expanding both sides: $18x^2 - 15x - 6x + 5 = 10x^2 - 6x + 5x - 3$. $18x^2 - 21x + 5 = 10x^2 - x - 3$. Rearranging the terms: $8x^2 - 20x + 8 = 0$. Dividing by $4$: $2x^2 - 5x + 2 = 0$. Factoring the quadratic equation: $2x^2 - 4x - x + 2 = 0 \implies 2x(x - 2) - 1(x - 2) = 0$. $(2x - 1)(x - 2) = 0$. Thus,$x = 0.5$ or $x = 2$. If $x = 0.5$,$PX = 3(0.5) - 1 = 0.5$ and $PW = 6(0.5) - 5 = -2$,which is impossible as length cannot be negative. Therefore,$x = 2$.

In $\square XYZW$,$\overline{XY} \parallel \overline{ZW}$ and $\overline{XZ} \cap \overline{YW} = \{P\}$. If $PX = 3x - 1$,$PY = 2x + 1$,$PZ = 5x - 3$,and $PW = 6x - 5$,find the value of $x$.

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