Question

PLZ HELP ME SOLVE THIS QUESTION THANKS​

Answer 1

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

Given that,

Following three points are the vertices of triangle,

• A (x, - 3)

• B(4, 6)

• C(6, - 3)

and

• Area of triangle ABC = 45 sq. units

We know,

☆ Area of triangle is given by

$\rm :\longmapsto\:\ Area =\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Here,

$\rm :\longmapsto\:Area = 45$

$\rm :\longmapsto\:x_1 = x$

$\rm :\longmapsto\:x_2 = 4$

$\rm :\longmapsto\:x_3 = 6$

$\rm :\longmapsto\:y_1 = - 3$

$\rm :\longmapsto\:y_2 = 6$

$\rm :\longmapsto\:y_3 = - 3$

Thus, on substituting the values, we get

$\rm :\longmapsto\: \pm \: 45 = \dfrac{1}{2} \bigg(x(6 + 3) + 4( - 3 + 3) + 6( - 3 - 6)\bigg)$

$\rm :\longmapsto\: \pm \: 45 = \dfrac{1}{2} \bigg(9x+ 6( - 9)\bigg)$

$\rm :\longmapsto\: \pm \: 45 = \dfrac{9}{2} \bigg(x - 6\bigg)$

$\rm :\longmapsto\: \pm \: 5 = \dfrac{1}{2} \bigg(x - 6\bigg)$

$\rm :\longmapsto\: \pm \: 10 = x - 6$

$\rm :\longmapsto\:10 = x - 6 \: \: \: or \: \: \: - 10 = x - 6$

$\rm :\longmapsto\:10 + 6 = x \: \: \: or \: \: \: - 10 + 6 = x$

$\rm :\longmapsto\:16 = x \: \: \: or \: \: \: - 4= x$

$\bf\implies \:x = 16 \: \: as \: x > 0$

Additional Information :-

1. Distance Formula :-

$\rm :\longmapsto \:AB = \sqrt{ {(x_2-x_1)}^{2} + {(y_2-y_1)}^{2} }$

2. Section Formula :-

$\rm :\longmapsto \:( x, y) = \bigg (\dfrac{mx_2 + nx_1}{m + n} , \dfrac{my_2 + ny_1}{m + n} \bigg)$

3. Midpoint Formula :-

$\rm :\longmapsto\:( x, y) = \bigg(\dfrac{x_1+x_2}{2} , \dfrac{y_1+y_2}{2} \bigg )$