The line L given by [x 5 + y b = 1] passes through the point [( 13,32 )].

Question

The line $L$ given by $${x \over 5} + {y \over b} = 1$$ passes through the point $$\left( {13,32} \right)$$. The line K is parrallel to $L$ and has the equation $${x \over c} + {y \over 3} = 1.$$ Then the distance between $L$ and $K$ is :

Accepted Answer

Slope of line $$L =  - {b \over 5}$$

Slope of line $$K =  - {3 \over c}$$

Line $L$ is parallel to line $k.$

$$ \Rightarrow {b \over 5} = {3 \over c} \Rightarrow bc = 15$$

$(13,32)$ is a point on $L.$

$$\therefore$$ $${{13} \over 5} + {{32} \over b} = 1 \Rightarrow {{32} \over b} =  - {8 \over 5}$$

$$ \Rightarrow b =  - 20 \Rightarrow c =  - {3 \over 4}$$

Equation of $K:$ $y - 4x = 3$

$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ \Rightarrow 4x - y + 3 = 0$$

Distance between $L$ and $K$

$$ = {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$

Solutions · 1