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Standard XII

Mathematics

Section Formula

Let h, k be a...

Question

Let (h, k) be a fixed point, where $h > 0, k > 0$ . A straight line passing through this point cuts the positive direction of the coordinates axes at the points P and Q. Find the minimum area of the triangle OPQ, O being the origin.

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Solution

let a and b be the intercepts made by the line on x and y axis.

equation of line will be $\frac{x}{a} + \frac{y}{b} = 1$

the area of triangle made by this line will be $\frac{1}{2} \times a \times b$

since this line passes through (h,k).put it in that equation.then find the relation between a,b,h,k.replace b in terms of a,h,k in area expression.

Then minimize it and find the value of a.

So min.area will occur when (h,k) is the middle point of the section of line intercepted between axis.in that case $a = 2 h, b = 2 k$ so min.area will be $= 2 h k$

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Let (h, k) be a fixed point, where h>0,k>0. A straight line passing through this point cuts the positive direction of the coordinates axes at the points P and Q. Find the minimum area of the triangle OPQ, O being the origin.

Let (h, k) be a fixed point, where $h > 0, k > 0$ . A straight line passing through this point cuts the positive direction of the coordinates axes at the points P and Q. Find the minimum area of the triangle OPQ, O being the origin.