Get 2 FREE Instant-Explanations on Filo with code FILOAPP

For what value of k(k > 0) is the area of the triangle with ve

Question

For what value of $k (k > 0)$ is the area of the triangle with vertices $(- 2, 5), (k, - 4)$ and $(2 k + 1, 10)$ equal to $53$ square units?

Views: 751 students

Found 7 tutors discussing this question

Scarlett Discussed

For what value of

k (k > 0)

is the area of the triangle with vertices

(- 2, 5), (k, - 4)

and

(2 k + 1, 10)

equal to

53

square units?

9 mins ago

Discuss this question LIVE

9 mins ago

Text solutionVerified

Area of triangle

= 53

square units

Area of

Δ A BC = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]

53 = \frac{1}{2} [- 2 (- 4 - 10) + k (10 - 5) + (2 k + 1) (5 + 4)]

53 = \frac{1}{2} [- 2 \times (- 14) + k \times 5 + (2 k + 1) \times 9]

106 = 23 k + 37

k = \frac{69}{23} = 3

The value of k is

3

Was this solution helpful?

150

Report

One destination to cover all your homework and assignment needs

Learn Practice Revision Succeed

Instant 1:1 help, 24x7

60, 000+ Expert tutors

Textbook solutions

Big idea maths, McGraw-Hill Education etc

Essay review

Get expert feedback on your essay

Schedule classes

High dosage tutoring from Dedicated 3 experts

Trusted by 4 million+ students

Stuck on the question or explanation?

Connect with our math tutors online and get step by step solution of this question.

231 students are taking LIVE classes

Question Text	For what value of $k (k > 0)$ is the area of the triangle with vertices $(- 2, 5), (k, - 4)$ and $(2 k + 1, 10)$ equal to $53$ square units?
Answer Type	Text solution:1
Upvotes	150