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

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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: $3a + 2b = 13$ ..... $(i)$ Given that point $Q(b, a)$ lies on $4x - y = 5$,we have: $4b - a = 5$ or $-a + 4b = 5$ ..... $(ii)$ Multiplying equation $(ii)$ by $3$,we get $-3a + 12b = 15$ ..... $(iii)$ Adding $(i)$ and $(iii)$: $(3a + 2b) + (-3a + 12b) = 13 + 15$ $14b = 28 \Rightarrow b = 2$ Substituting $b = 2$ in $(i)$: $3a + 2(2) = 13$ $\Rightarrow 3a = 9$ $\Rightarrow a = 3$ 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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