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coordinate

by daretodream » Fri Feb 19, 2010 3:25 am
The co-ordinates of the foot of the perpendicular from the point (2, 2) on the line y - 2x - 8 = 0 are

(A) (-1, 6)

(B) (-1, -4)

(C) (-2, 10)

(D) (2, 4)

(E) (-2, 4)
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by harsh.champ » Fri Feb 19, 2010 4:06 am
daretodream wrote:The co-ordinates of the foot of the perpendicular from the point (2, 2) on the line y - 2x - 8 = 0 are

(A) (-1, 6)

(B) (-1, -4)

(C) (-2, 10)

(D) (2, 4)

(E) (-2, 4)
The line is y=2x + 8 [m1=2 (m denotes the slope of the line)]
If a line is perpendicular to another line ,then we have m1 x m2 =-1
Hence,m2 = -1/2(Slope of line passing through 2,2)
y=mx + c
putting (2,2) and m=-1/2,we have c=3
y=(-1/2)x + 3

SO the co-ordinate will be the intersecton of the 2 lines.
It is similar to solving 2 eqns. and 2 variables.

Putting y=2x + 8 in the 2nd eqn.,we have
2x+8 = (-1/2)x + 3
Thus,x=-2 .
So,y=4

So, the answer [spoiler]will be (-2,4) E.[/spoiler]
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by vijay_venky » Fri Feb 19, 2010 6:05 am
The method discussed by harsh.champ is good and procedural.

But I think we can proceed with the answer choices as well.

First the foot of the perpendicular should lie on the line. So check which points are on the line y=2(x+4)
Only two points obey this A and E.

And this point should lie on the perpendicular so the product of the slope of the line connecting this and (2,2) with 2(slope of y=2x+8) should be -1. This is satisfied by only E.
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by harsh.champ » Fri Feb 19, 2010 6:12 am
vijay_venky wrote:The method discussed by harsh.champ is good and procedural.

But I think we can proceed with the answer choices as well.

First the foot of the perpendicular should lie on the line. So check which points are on the line y=2(x+4)
Only two points obey this A and E.

And this point should lie on the perpendicular so the product of the slope of the line connecting this and (2,2) with 2(slope of y=2x+8) should be -1. This is satisfied by only E.
Hey venky,
What I think is that the method of plugging equations takes quite some time whereas my soln. approach, though looks lengthy but when you actually calculate it it can be done within a minute.

In case a person forgets that m1 x m2 = -1,I do admit that your approach can come in handy.
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



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by shashank.ism » Fri Feb 19, 2010 6:43 am
daretodream wrote:The co-ordinates of the foot of the perpendicular from the point (2, 2) on the line y - 2x - 8 = 0 are

(A) (-1, 6)

(B) (-1, -4)

(C) (-2, 10)

(D) (2, 4)

(E) (-2, 4)
the line is y=2x+8-------------(i) its slope = m=2
slope of line perpendicular to it = m' =-1/2
so eq of line is say y= -1/2 x +c since (2,2) lies on this line so 2=-1/2 x 2+c --> c=3
so y= -1/2 x +3 ---------(ii)
solving (i) and (ii) , we get
2x+8 = -1/2 x +3 --> 4x+16 = -x+6 5x=-10 x=-2
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by shashank.ism » Fri Feb 19, 2010 6:45 am
vijay_venky wrote:The method discussed by harsh.champ is good and procedural.

But I think we can proceed with the answer choices as well.

First the foot of the perpendicular should lie on the line. So check which points are on the line y=2(x+4)
Only two points obey this A and E.

And this point should lie on the perpendicular so the product of the slope of the line connecting this and (2,2) with 2(slope of y=2x+8) should be -1. This is satisfied by only E.
Venky I think your approach is also a good one..since only two options are matching when we plug in ...so we can check the slope...Its also quite fast....
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