Question

If the vertices of the triangle are $\left( 1,k \right),\left( 4,-3 \right),\left( -9,7 \right)$ and its area is 15 square units, find the value of k.

Answer 1

Hint: Use the area formula which is,
Area $=\dfrac{1}{2}\left| {{a}_{1}}\left( {{b}_{2}}-{{b}_{3}} \right)+{{a}_{2}}\left( {{b}_{3}}-{{b}_{1}} \right)+{{a}_{3}}\left( {{b}_{1}}-{{b}_{2}} \right) \right|$
Where $\left( {{a}_{1}},{{b}_{1}} \right)\left( {{a}_{2}},{{b}_{2}} \right)\left( {{a}_{3}},{{b}_{3}} \right)$ are points and the equate it with 15.

Complete step-by-step answer:

In the question, we are given vertices of triangle a co-ordinates which are $\left( 1,k \right),\left( 4,-3 \right),\left( -9,7 \right)$ and its area is 15 sq. units is also given we have to find the value of k.
For finding the value of k we will use the formula of area which is,
Area $=\dfrac{1}{2}\left| {{a}_{1}}\left( {{b}_{2}}-{{b}_{3}} \right)+{{a}_{2}}\left( {{b}_{3}}-{{b}_{1}} \right)+{{a}_{3}}\left( {{b}_{1}}-{{b}_{2}} \right) \right|$
Where points of vertices are $\left( {{a}_{1}},{{b}_{1}} \right)\left( {{a}_{2}},{{b}_{2}} \right)\left( {{a}_{3}},{{b}_{3}} \right)$ of a triangle whose area to be find out.
Here, points are $\left( 1,k \right),\left( 4,-3 \right),\left( -9,7 \right)$ so, area will be
Area $=\dfrac{1}{2}\left| 1\left( -3-7 \right)+4\left( 7-k \right)-9\left( k+3 \right) \right|$
Which can further simplified as
Area $=\dfrac{1}{2}\left| -10+28-4k-9k-27 \right|$
$=\dfrac{1}{2}\left| -13k-9 \right|$
We know the area of the triangle is 15 sq. units.
$=\dfrac{1}{2}\left| -13k-9 \right|=15$
On cross multiplication, we get
$=\left| -13k-9 \right|=30$
When $\left| x \right|=a$, where a is constant then removing after removing mod sign we get $x=\pm a$.
So, we can write as
$-13k-9=\pm 30$
On simplification, we get
$-13k=9\pm 30$
So,
$-13k=39$ or \[-21\]
Hence, \[k=\dfrac{-39}{13}\] or \[\dfrac{21}{13}\]
Hence, \[k=-3\] or \[\dfrac{21}{13}\].

Note: One can also use distance formula to find distance between all the points and then use Heron’s formula and equate it with the given area but this process will be very lengthy and tedious.