JEE MAIN - Mathematics (2010 - No. 17)
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 :
$$\sqrt {17} $$
$${{17} \over {\sqrt {15} }}$$
$${{23} \over {\sqrt {17} }}$$
$${{23} \over {\sqrt {15} }}$$
Explanation
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 $$K$$
$$ = {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$
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 $$K$$
$$ = {{\left| {52 - 32 + 3} \right|} \over {\sqrt {17} }} = {{23} \over {\sqrt {17} }}$$
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