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Mathematics - Straight Lines Question with Solution | TestHub

MathematicsStraight LinesIntersection of linesMedium2 minQB

MathematicsMediuminteger

In the first quadrant, let $A$ be the area bounded by the $y$ -axis, above by $7 x + 4 y = 168$ , and below by $5 x + 3 y = 121$ . Find $3 A /10$ .

Answer:

Solution:

The lines meet at $(20, 7)$ and cross the $y$ -axis at heights 42 and $121/3$ . Their triangular area is $A = \frac{1}{2} (42 -$ $121/3) (20) = 50/3$ , hence $3 A /10 = 5$ .

Stream:JEESubject:MathematicsTopic:Straight LinesSubtopic:Intersection of lines

⏱ 2mℹ️ Source: QB

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