JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 16)
A straight line through origin O meets the lines 3y = 10 − 4x and 8x + 6y + 5 = 0 at points A and B respectively. Then O divides the segment AB in the ratio :
2 : 3
1 : 2
4 : 1
3 : 4
Explanation
The lines 4x + 3y $$-$$ 10 = 0 and
8x + 6y + 5 = 0 , are parallel as
$${4 \over 8}$$ = $${3 \over 6}$$
Now length of perpendicular from
(0, 0, 0) to 4x + 3y $$-$$ 10 = 0 is,
P1 = $$\left| {{{4\left( 0 \right) + 3\left( 0 \right) - 10} \over {\sqrt {{4^2} + {3^2}} }}} \right|$$ = $${{10} \over 5}$$ = 2
Length of perpendicular from
0 (0, 0) to 8x + 6y + 5 = 0 is
P2 = $$\left| {{{8\left( 0 \right) + 6\left( 0 \right) + 5} \over {\sqrt {{6^2} + {8^2}} }}} \right|$$ = $${5 \over {10}}$$ = $${1 \over 2}$$
$$\therefore\,\,\,$$ P1 : P2 = 2 : $${1 \over 2}$$ = 4 : 1
8x + 6y + 5 = 0 , are parallel as
$${4 \over 8}$$ = $${3 \over 6}$$
Now length of perpendicular from
(0, 0, 0) to 4x + 3y $$-$$ 10 = 0 is,
P1 = $$\left| {{{4\left( 0 \right) + 3\left( 0 \right) - 10} \over {\sqrt {{4^2} + {3^2}} }}} \right|$$ = $${{10} \over 5}$$ = 2
Length of perpendicular from
0 (0, 0) to 8x + 6y + 5 = 0 is
P2 = $$\left| {{{8\left( 0 \right) + 6\left( 0 \right) + 5} \over {\sqrt {{6^2} + {8^2}} }}} \right|$$ = $${5 \over {10}}$$ = $${1 \over 2}$$
$$\therefore\,\,\,$$ P1 : P2 = 2 : $${1 \over 2}$$ = 4 : 1
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