Question

If the area of the triangle formed by the vertices (-3,5)(5,-2) and (5,a).(take in order) 32 sq units. find the value of a​

Answer 1

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

Given that

The vertices of triangle is given as ( - 3, 5 ), ( 5, - 2 ) and ( 5, a )

and Area of triangle = 32 square units.

We know, area of triangle is given by

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

Now, we have

• • x₁ = - 3

• • x₂ = 5

• • x₃ = 5

• • y₁ = 5

• • y₂ = - 2

• • y₃ = a

And

• Area = 32

So, on substituting all these values in above formula, we get

$\rm :\longmapsto\:\sf 32 =\dfrac{1}{2}\bigg| - 3( - 2 - a) + 5(a - 5) + 5(5 + 2)\bigg |$

$\rm :\longmapsto\:\sf 64 =\bigg| 6 + 3a + 5a - 25 + 25 + 10\bigg |$

$\rm :\longmapsto\:\sf 64 =\bigg| 16 + 8a\bigg |$

$\rm :\longmapsto\:16 + 8a = \: \pm \: 64$

$\rm :\longmapsto\:16 + 8a = 64 \: \: or \: \: 16 + 8a = - 64$

$\rm :\longmapsto\:8a = 64 - 16\: \: or \: \:8a = - 64 - 16$

$\rm :\longmapsto\:8a =48\: \: or \: \:8a = -80$

$\bf\implies \:a = 6 \: \: or \: \: a = - \: 10$

Additional Information :-

1. Distance Formula :-

${\underline{\boxed{\frak{\quad Distance = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \quad}}}}$

2. Section Formula :-

${\underline{\boxed{\frak{\quad \Big(x, y \Big) = \Bigg(\dfrac{mx_2 + nx_1}{m + n},  \dfrac{my_2 + ny_1}{m + n}\Bigg) \quad}}}}$

3. Midpoint Formula :-

${\underline{\boxed{\frak{\quad (x, \: y) \: = \: \bigg(\dfrac{x_1 + x_2}{2} \; ,\; \dfrac{y_1 + y_2}{2} \bigg)\quad}}}}$