Question

Find the value of k so that the area of the triangle with vertices A (k+1, 1), B(4, -3) and C(7, -k) is 6 square units.

Answer 1

Answer:

Step-by-step explanation:

Given :-

The vertices are (k + 1, 1) (4, -3) and (7, -k) and the area of the triangle is 6 square units.

To Find :-

The value of K.

Formula to be used :-

$\boxed {\underline{Area=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]}}$

Solution:-

Putting all the values, we get

$\implies 6=\frac{1}{2}[(k+1)(-3+k)+4(-k-1)+7(1+3)]$

$\implies 12=(k+1)(k-3)+4(-k-1)+28$

$\implies 12=k^2-3k+k-3-4k-4+28$

$\implies k^2-6k+9=0$

$\implies k^2-3k-3k+9=0$

$\implies k(k-3)-3(k-3)=0$

$\implies (k-3)(k-3)=0$

$\implies k=3,3$

Hence, The value of k is 3.

Answer 2

Solution:

Given:

=> Area of triangle = 6 units.

=> Vertices A = (k + 1, 1)

=> Vertices B = (4, -3)

=> Vertices C = (7, -k)

To Find:

=> Value of k.

Formula used:

$\sf{\implies Area = \dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]}$

Now, put the values in the formula.

$\sf{\implies Area = \dfrac{1}{2}[x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})]}$

$\sf{\implies 6=\dfrac{1}{2}[(k+1)(-3+k) +4(-k-1)+7(1+3)]}$

$\sf{\implies 12=[-3k+k^{2}-3+k-4k-4+7+21]}$

$\sf{\implies 12 = k^{2}-6k+21}$

$\sf{\implies k^{2}-6k+21-12=0}$

$\sf{\implies k^{2}-6k+9=0}$

We can solve it by splitting middle term method.

$\sf{\implies k^{2}-6k+9=0}$

$\sf{\implies k^{2}-3k-3k+9=0}$

$\sf{\implies k(k-3)-3(k-3)=0}$

$\sf{\implies (k-3)(k-3)}$

$\sf{\implies k = 3,3}$

Hence, The value of k is 3.