sanju09 wrote:The area of a triangle is 21. If two of its vertices lie on points (5, 3) and (-4, -3), then which of the following represents the equation of line that could contain the third vertex of triangle?
(A) 2 x - 3 y = 15
(B) 2 x + 3 y = 15
(C) 3 x + 2 y = 15
(D) 3 x - 2 y = 15
(E) both (B) and (D)

Even if the line from dat pt. is the altitude we will get the point.
(1,0) will be the mid-pt.
Now,area = 21
My method is a long one:-Calculating from Hero's formula.
Any short method??

You can get the eqn of line which join vertices points (5, 3) and (-4, -3)
-->y-3/x-5 = 6/9 ==> 2x - 3y - 1 = 0

Now the third verex got to be in some line perpendicular to the line joining points (5, 3) and (-4, -3)

When you calculate area you take base *height/2 Here base you can take as line joining points (5, 3) and (-4, -3) and height a perpendicular dropped from the third vertex. But in this particular case you dont really need to worry about actual area size , etc.

Eqn of line that can be dropped from third vertex perpendicular to base is of form 3x + 2y + c = 0,

Now looking at the choices one choice is of this form , i,e choice C .

harsh.champ wrote:

sanju09 wrote:The area of a triangle is 21. If two of its vertices lie on points (5, 3) and (-4, -3), then which of the following represents the equation of line that could contain the third vertex of triangle?
(A) 2 x - 3 y = 15
(B) 2 x + 3 y = 15
(C) 3 x + 2 y = 15
(D) 3 x - 2 y = 15
(E) both (B) and (D)

Even if the line from dat pt. is the altitude we will get the point.
(1,0) will be the mid-pt.
Now,area = 21
My method is a long one:-Calculating from Hero's formula.
Any short method??

A line parallel to the line segment joining the points (5, 3) and (-4, -3), and at a distance of the calculable altitude, would contain all points those are needful here. We really don't need to do that hard work, though.

rohan_vus wrote:You can get the eqn of line which join vertices points (5, 3) and (-4, -3)
-->y-3/x-5 = 6/9 ==> 2x - 3y - 1 = 0

Now the third verex got to be in some line perpendicular to the line joining points (5, 3) and (-4, -3)

When you calculate area you take base *height/2 Here base you can take as line joining points (5, 3) and (-4, -3) and height a perpendicular dropped from the third vertex. But in this particular case you dont really need to worry about actual area size , etc.

Eqn of line that can be dropped from third vertex perpendicular to base is of form 3x + 2y + c = 0,

Now looking at the choices only one choice is of this form , i,e choice C

Why is it compulsory that the line will be perpendicular ??
Its not given that AD will be necessarily a altitude.

Surely, any line that doesn't go through both of the vertices already given could create a triangle with any area?

harsh.champ wrote:

rohan_vus wrote:You can get the eqn of line which join vertices points (5, 3) and (-4, -3)
-->y-3/x-5 = 6/9 ==> 2x - 3y - 1 = 0

Now the third verex got to be in some line perpendicular to the line joining points (5, 3) and (-4, -3)

When you calculate area you take base *height/2 Here base you can take as line joining points (5, 3) and (-4, -3) and height a perpendicular dropped from the third vertex. But in this particular case you dont really need to worry about actual area size , etc.

Eqn of line that can be dropped from third vertex perpendicular to base is of form 3x + 2y + c = 0,

Now looking at the choices only one choice is of this form , i,e choice C

Why is it compulsory that the line will be perpendicular ??
Its not given that AD will be necessarily a altitude.

Its not about being compulsory . Its about possibility . Infact any line could be part of the third vertex provided it gives you calculabe area.
Ok , lets , go step further

The lenght of base = sqrt((5+4)^2 +(3+3)^2) = sqrt(117) = 3*sqrt(13) , so length of altitude for the triangle could be 14/sqrt(13)

Let say vertex is (a,b) ..The height of the triangle w.r.t its base would be the perpendicular distance from (a,b) to the base.. perpendicular distance to a line 2x-3y-1 = |2a-3b-1|/sqrt(13) , this gives |2a-3b-1| = 14

This eqn is satisfied by choice A (2x-3y-15= 0) . A vertex can be anywhere on this line , it doesnt matter .
But as far as eqn of third vetex goes and question says ( what equation of line that could contain the third vertex ) then even choice C satisfies that

Choice C satisfies eqn ( 3 x + 2 y = 15) for value of X = 75/13 and y = -15/13 , this co-ordinate also gives a perprndicular distance of 14/sqrrt(13) from the base

So there are more than one correct answer here. Question , if would have stated , "eqn of third vertex must always be part of which eqn", then A would have been the answer .