The point P(a, b) lies on the straight line 3 | vedclass

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(B) Given that point $P(a, b)$ lies on $3x + 2y = 13$,we have $3a + 2b = 13$ (Equation $1$).Given that point $Q(b, a)$ lies on $4x - y = 5$,we have $4

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$x + y = 5$

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(B) Given that point $P(a, b)$ lies on $3x + 2y = 13$,we have $3a + 2b = 13$ (Equation $1$).Given that point $Q(b, a)$ lies on $4x - y = 5$,we have $4b - a = 5$,which implies $a = 4b - 5$ (Equation $2$).Substituting Equation $2$ into Equation $1$:$3(4b - 5) + 2b = 13$$12b - 15 + 2b = 13$$14b = 28 \implies b = 2$.Substituting $b = 2$ into Equation $2$:$a = 4(2) - 5 = 8 - 5 = 3$.Thus,$P = (3, 2)$ and $Q = (2, 3)$.The slope $m$ of line $PQ$ is $\frac{3 - 2}{2 - 3} = \frac{1}{-1} = -1$.The equation of line $PQ$ passing through $(3, 2)$ is:$y - 2 = -1(x - 3)$$y - 2 = -x + 3$$x + y = 5$.

The point $P(a, b)$ lies on the straight line $3x + 2y = 13$ and the point $Q(b, a)$ lies on the straight line $4x - y = 5$. Then the equation of the line $PQ$ is

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