Question

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

Answer 1

Hint: We are given with the vertices of the triangle. We will solve this problem by putting the coordinates in the formula \[\dfrac{1}{2}\left[ {{a_x}({b_y} - {c_y}) + {b_x}({c_y} - {a_y}) + {c_x}({a_y} - {b_y})} \right]\] where $({a_x}, {a_y}), ({b_x}, {b_y}), ({c_x}, {c_y})$ are the coordinates of vertices of the triangle. Thus we will get a quadratic equation in terms of p. By solving that using middle term factorization or putting quadratic formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ we will get the values of p.

Complete step-by-step answer:
Let A(1,-1), B(-4,2k) and C(-k,-5) be the vertices of a triangle.
Hence the area of the triangle ABC is,
\[\dfrac{1}{2}\left[ {{a_x}({b_y} - {c_y}) + {b_x}({c_y} - {a_y}) + {c_x}({a_y} - {b_y})} \right]\]
Putting coordinates A, B, C in the above formula we get,
\[\dfrac{1}{2}\left[ {1(2k - ( - 5)) + ( - 4)(( - 5) - ( - 1)) + ( - k)( - 1 - 2k)} \right]\]
Simplifying it we get,
\[\dfrac{1}{2}\left[ {(2k + 5) + ( - 4)( - 5 + 1)) + ( - k)( - 1 - 2k)} \right]\]
\[ = \dfrac{1}{2}\left[ {2k + 5 + 20 - 4 + k + 2{k^2}} \right]\]
\[ = \dfrac{1}{2}\left[ {2{k^2} + 3k + 21} \right]\]………..(1)
But we have the area of the triangle is 24 square units.
Comparing it with above equation,
\[\dfrac{1}{2}\left[ {2{k^2} + 3k + 21} \right] = 24\]
By cross multiplication we get,
\[2{k^2} + 3k + 21 = 48\]
Taking 48 to the left hand side we get,
\[2{k^2} + 3k + 21 - 48 = 0\]
$ \Rightarrow 2{k^2} + 3k - 27 = 0$
By doing middle term factorization we get,
$2{k^2} + 9k - 6k - 27 = 0$
Taking k common from first two terms and -3 from second two terms we get,
$k(2k + 9) - 3(2k + 9) = 0$
$ \Rightarrow (2k + 9)(k - 3) = 0$
Taking $2k + 9 = 0$ we get $k = \dfrac{-9}{2}$
Again taking $k - 3 = 0$ we get $k = 3$
Hence the values of k are 3 and -9/2.

Note: If the vertices of a triangle be $({a_x},{a_y}),({b_x},{b_y}),({c_x},{c_y})$, then the area of that triangle is given by \[\dfrac{1}{2}\left[ {{a_x}({b_y} - {c_y}) + {b_x}({c_y} - {a_y}) + {c_x}({a_y} - {b_y})} \right]\]
We can also solve the same problem using alternative methods.
Alternative method- You can get the sides of the triangle using formula $\sqrt {(x{'^2} - {x^2}) + (y{'^2} - {y^2})} $from the coordinates of vertices. Then by using formula $\sqrt {S(S - a)(S - b)(S - c)} $you will get the area of the triangle equation where S is the half of the perimeter of triangle and a, b, c are the lengths of the sides of the triangle. Thus you will get the quadratic equation and by solving that you will get the values of k.
You should practice more questions from coordinate geometry.
You should remember all the formulas of coordinate geometry.
To solve the quadratic equation we use two methods. One is quadratic formula of obtaining roots i.e. $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and another is middle term factorization in which we split coefficient of middle term in such a way that its product is equal to the last term.
You should practice more and more quadratic equations using these two methods.