Question

a line passing through a point a 2 4 intersects the line X + Y is equals to 9 at point b what is the minimum distance between a b​

Answer 1

Question :-

A line passing through a point A(2,4) , intersect the line x + y = 9 at point B then what is the minimum distance between A and B .

Answer :-

$\to \boxed{ \sf{ r = \frac{3}{ \sqrt{2} } }} \bf{ units \: }$

To Find :-

→ Minimum distance between A and B .

Explanation :-

We will solve it by parametric equation .

We know that ,

$\to \boxed{ \sf{ \frac{x - x_{1} }{ \cos \theta} = \frac{y -y_{1} }{ \sin \theta} = r \: }}$

here "r" is the distance of a point on a pine from a point .

Therefore ,it passes through (2,4)

$\to \sf{ \frac{x - 2 }{ \cos \theta} = \frac{y -4 }{ \sin \theta} = r \: } \\ \: \\ \therefore \\ \implies \sf{ \: x = r \cos \theta + 2 \: } \\ \bf and \: \\ \implies \sf{ y = r \sin \theta \: + 4}$

it interesting on x + y = 9

hence,

$\to \sf{r \cos \theta + 2 + r \sin \theta + 4 = 9} \\ \\ \to \sf{ r( \sin \theta \: + \cos \theta) = 9 - 6} \\ \\ \to \sf{ r = \frac{3}{ \sin \theta + \cos \theta} }$

slope of line x + y = 9 is -1

hence ,

$\to \: \tan \theta = - 1 \: \\ \\ \to \theta = 135 \degree \\\sf{ \:but\:angle\:make\:by\:other\:line\:is\:45 }\\ \\ \to \sf{ r = \frac{3}{ \sin(45) + \cos(45)} } \\ \\ \to \sf{ r = \frac{3}{ \frac{1}{ \sqrt{2}} + \frac{1}{ \sqrt{2} }}} \\ \\ \to \sf{r = \frac{3}{ \frac{2}{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } } } \\ \\ \to \boxed{ \sf{ r = \frac{3}{ \sqrt{2} } }}$

hence minimum distance is 3/√2 units .