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(B) To find the point of intersection of the lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$,we solve them simultaneously.Multiply the first equation by

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$5x + 4y = 0$

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(B) To find the point of intersection of the lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$,we solve them simultaneously.Multiply the first equation by $2$: $2x + 4y - 20 = 0$.Subtract the second equation from this: $(2x + 4y - 20) - (2x + y + 5) = 0$,which simplifies to $3y - 25 = 0$,so $y = 25/3$.Substitute $y = 25/3$ into $x + 2y - 10 = 0$: $x + 2(25/3) - 10 = 0 \implies x + 50/3 - 30/3 = 0 \implies x = -20/3$.The point of intersection is $(-20/3, 25/3)$.Testing the options,for $5x + 4y = 0$:$5(-20/3) + 4(25/3) = -100/3 + 100/3 = 0$.Thus,the line $5x + 4y = 0$ passes through the point of intersection.

The straight line that passes through the point of intersection of the straight lines $x + 2y - 10 = 0$ and $2x + y + 5 = 0$ is:

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