Math, asked by kunalrajbth02pa2en7, 11 months ago

find the value of k for so that the area of the triangle with vertices (1,-1),(-4,2k) and (-k,-5)is 24 square units.

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Answered by pranamsinha

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kunalrajbth02pa2en7: no it is wrong answer

Anonymous: Two answers. Either k = 3 or k = -4.5.

kunalrajbth02pa2en7: not k=3

Anonymous: Without doing any maths, you can do this by putting the three points into Geogebra; for the two points depending on k, make a slider for the k variable. When you make a triangle with the polygon tool, selecting the three points, Geogebra shows the area of that triangle. Now just slide through different values of k and see which ones give the necessary area.

Anonymous: Otherwise, there is a nice little rule involving a 3x3 determinant for the area of a triangle in terms of the coordinates of the vertices. Using this rule, we get that plus or minus double the area is given by 2k^2 + 3k + 21, so this has to be equal to either -48 or 48 (that's plus or minus double the area). When we put -48, there are no solutions to this quadratic. When we put 48, the quadratic factorizes as (2k + 9)(k - 3) = 0. Hence the two solutions 3 and -4.5.

Anonymous: Yes k=3 is a solution! Then the vertices are at (1, -1), (-4, 6) and (-3, -5), and the area is 24, as required.

kunalrajbth02pa2en7: you don't use discriminant method for this question

Anonymous: That's a quick and easy way to get the area when the coordinates of the vertices are known. Since here we have the coordinates in terms of k, this is a quick and easy way to get the area in terms of k, giving an equation to solve.

Anonymous: Sure, you could go via a more tedious route, but of course, the answer would be the same. Again, I would encourage you to check the answer in software like Geogebra. You could also confirm with wolframalpha. Just type in something like "area of triangle with vertices (1,-1) (-4,6) (-3,-5)"

Anonymous: Since an answer got deleted, I've put in the explanation from these comments as a solution. Happy problem solving!

Answered by Anonymous

Answer:

There are two values of k that meet the requirement: k = 3 and k = -4.5

Step-by-step explanation:

Without doing any maths, you can do this by putting the three points into Geogebra. For the two points depending on k, make a slider for the k variable. When you make a triangle with the polygon tool, selecting the three points, Geogebra shows the area of that triangle. Now just slide through different values of k and see which ones give the necessary area. See the attachments.

Otherwise, there is a nice little rule involving a 3x3 determinant for the area of a triangle in terms of the coordinates of the vertices:

>> Write 1's down the first column, then in the 6 remaining spaces, put in the coordinates of the points (x coordinates down 2nd column and y coordinates down the 3rd column). The determinant's value is ±2 times the area of the triangle.

Using this rule, we get that plus or minus double the area is given by

det ( 1, 1, -1 ; 1, -4, 2k ; 1, -k -5 ) = 2k² + 3k + 21,

so this has to be equal to either -48 or 48 (that's plus or minus double the area).

When we put -48, there are no solutions to this quadratic (the discriminant is 3² - (4)(2)(69), and this is negative).

When we put 48, the quadratic is

2k² + 3k + 21 = 48

=> 2k² + 3k - 27 = 0

=> (2k + 9)(k - 3) = 0

=> k = 3 or k = -4.5.

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