If 12a + 5b = 9,where a, b in R,then the mini | vedclass

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(A) Given the linear equation $12a + 5b = 9$.We want to minimize $f(a, b) = a^2 + b^2$,which represents the square of the distance from the origin $(0

Vedclass · Accepted Answer

$\frac{81}{169}$

Vedclass · Answer

(A) Given the linear equation $12a + 5b = 9$.We want to minimize $f(a, b) = a^2 + b^2$,which represents the square of the distance from the origin $(0, 0)$ to a point $(a, b)$ on the line $12a + 5b = 9$.The minimum distance from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.Here,$A = 12$,$B = 5$,$C = -9$,and $(x_0, y_0) = (0, 0)$.Thus,the minimum distance $d = \frac{|12(0) + 5(0) - 9|}{\sqrt{12^2 + 5^2}} = \frac{|-9|}{\sqrt{144 + 25}} = \frac{9}{\sqrt{169}} = \frac{9}{13}$.The minimum value of $a^2 + b^2$ is the square of this minimum distance $d^2 = (\frac{9}{13})^2 = \frac{81}{169}$.

If $12a + 5b = 9$,where $a, b \in R$,then the minimum value of $a^2 + b^2$ is -

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