JEE MAIN - Mathematics (2019 - 8th April Evening Slot - No. 20)
Suppose that the points (h,k), (1,2) and (–3,4) lie
on the line L1
. If a line L2
passing through the points
(h,k) and (4,3) is perpendicular to L1
, then
$$k \over h$$
equals :
$${1 \over 3}$$
3
0
-$${1 \over 7}$$
Explanation
Equation of line L1 passing through points (1, 2) and (–3, 4) is :
y - 2 = $$\left( {{{4 - 2} \over { - 3 - 1}}} \right)$$(x - 1)
$$ \Rightarrow $$ y - 2 = $$ - {1 \over 2}$$(x - 1)
$$ \Rightarrow $$ 2y – 4 = –x + 1
$$ \Rightarrow $$ x + 2y = 5 .......(L1)
Line L1 passes through point (h, k).
$$ \therefore $$ h + 2k = 5 .......(1)
Line L2 is perpendicular to line L1.
$$ \therefore $$ Equation of line L2 :
2x - y = $$\lambda $$
This line L2 passes through point (4, 3).
$$ \therefore $$ 2(4) - 3 = $$\lambda $$
$$ \Rightarrow $$ $$\lambda $$ = 5
$$ \therefore $$ Line L2 :
2x - y = 5
L2 also passes through (h, k)
$$ \therefore $$ 2h - k = 5 ............(2)
Solving (1) and (2) we get,
h = 3 and k = 1
$$ \therefore $$ $${k \over h} = {1 \over 3}$$
y - 2 = $$\left( {{{4 - 2} \over { - 3 - 1}}} \right)$$(x - 1)
$$ \Rightarrow $$ y - 2 = $$ - {1 \over 2}$$(x - 1)
$$ \Rightarrow $$ 2y – 4 = –x + 1
$$ \Rightarrow $$ x + 2y = 5 .......(L1)
Line L1 passes through point (h, k).
$$ \therefore $$ h + 2k = 5 .......(1)
Line L2 is perpendicular to line L1.
$$ \therefore $$ Equation of line L2 :
2x - y = $$\lambda $$
This line L2 passes through point (4, 3).
$$ \therefore $$ 2(4) - 3 = $$\lambda $$
$$ \Rightarrow $$ $$\lambda $$ = 5
$$ \therefore $$ Line L2 :
2x - y = 5
L2 also passes through (h, k)
$$ \therefore $$ 2h - k = 5 ............(2)
Solving (1) and (2) we get,
h = 3 and k = 1
$$ \therefore $$ $${k \over h} = {1 \over 3}$$
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