Question

if the area of the triangle formed by the vertices(-2,4)(3,-1)(5,k)is 10 square units find the value of k​

Answer 1

Answer:

9/5

Step-by-step explanation:

use determinant to solve it

Answer 2

The values of k are 1,-7

• Given vertices of triangle are

                      $A(x_{1},y_{1})=(-2,4)$

                      $B(x_{2},y_{2})=(3,-1)$

                      $C(x_{3},y_{3})=(5,k)$

• Area of triangle ABC is given by

                   $\frac{1}{2}|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|=10$

                   $\frac{1}{2}|-2(-1-k)+3(k-4)+5(4+1)|=10$

                   $\frac{1}{2}|2+2k+3k-12+25|=10$

                     5k + 15 = ±20

                   Case (i):

                     5k + 15 = 20

                     5k = 20 - 15

                     5k = 5

                       k = 1

                    Case (ii):

                     5k + 15 = -20

                     5k = -20 - 15

                     5k = -35

                       k = -7

 ∴ The values of k are 1,-7.