Question

2 points
2.
The area of ∆ABC is 32 Sq units. Where A(-5.-3).
B(-9. y), C(7.-8) then positive value of y is​

Answer 1

Answer:

Step-by-step explanation:

area=sqrt X1(y2-y3)+X2(y1-y3)+X3(y1-y2)

32=sqrt -5(y+8)-9(-8+3)+7(-3-y)

32=sqrt -5y-40+72-27=21-7y

32=sqrt -12y-5

Answer 2

$\large\underline{\bold{Given \:Question - }}$

• The area of ∆ ABC is 32 square units, where coordinates are A(- 5, - 3), B(-9, y) and C(7, - 8), then positive value of y is _______

$\large\underline{\bold{Solution-}}$

Concept Used :-

Area of triangle when vertices are given:-

$\sf \: Let \: us \: consider \: a \: triangle \: having \: vertices \: (x_1,y_1) \: , \: (x_2,y_2)$

$\sf \: and \: (x_3,y_3) \: respectively, \: then \: area \: of \: \triangle \: is \: given \: by \:$

$\sf \ Area =\dfrac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

Let's now solve the problem!!

Given

• A triangle ABC having area 32 square units with vertices A(-5, -3), B(-9, y) and C(7, -8).

So,

• we have

$\begin{gathered}\begin{gathered} \sf {(x_{1} , y_{1} )=( - 5, - 3)}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \sf {(x_{2} , y_{2} )=( - 9,y)}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered} \sf {(x_{3} , y_{3} )=(7, - 8)}\end{gathered}\end{gathered}$

Now,

We know,

• Area of triangle ABC is given by

$\sf \ Area =\dfrac{1}{2} |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) |$

On substituting the values, we get

$\sf \: 32 = \dfrac{1}{2} | - 5(y + 8) - 9( - 8 + 3) + 7( - 3 - y)|$

$\sf \: 64 = | - 5y - 40 + 45 - 21 - 7y|$

$\sf \: 64 = | - 12y - 16|$

$\sf \: 64 = |12y + 16|$

$\sf \: 12y + 16 = 64 \: \: \: or \: \: \: 12y + 16 = - 64$

$\sf \: 12y = 48 \: \: \: or \: \: \: 12y = - 64 - 16 \: \: \: \: (rejected \: as \: y \: > 0)$

$\: \boxed{ \bf{y = 4}}$

Additional Information :-

Condition for 3 points to be collinear.

Let us consider three vertices (a, b), (c, d) and (e, f), then these points are collinear or lies on a line if and only if

$\sf \: area_{(area \: of \: \triangle \: ABC)} = 0$