Question

What is the area of the triangle whose vertices are X(−5, −1)) , Y(−5, −10)) , and Z(−9, −7) ?

Answer 1

XY = −1−(−10) = 9 
(line x = −5) = −5−(−9) = 4

Area = 1/2 * 9 * 4 = 18

Answer 2

Answer:

17.96 units squared

Step-by-step explanation:

Refer the attached figure .

Point X =  (−5, −1)

Point Y=(−5, −10)

Point Z=(−9, −7)

In ΔXYZ , to find the length of sides we will use distance formula.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Length of XY

Point X = $(x_1,y_1)=(-5,-1)$

Point Y= $(x_2,y_2)=(-5,-10)$

$XY=\sqrt{(-5-(-5))^2+(-10-(-1))^2}$

$XY=\sqrt{(0)^2+(-9)^2}$

$XY=\sqrt{81}$

$XY=9$

Length of YZ

Point Y= $(x_1,y_1)=(-5,-10)$

Point Z = $(x_2,y_2)=(-9,-7)$

$YZ=\sqrt{(-9-(-5))^2+(-7-(-10))^2}$

$XY=\sqrt{(-4)^2+(3)^2}$

$YZ=\sqrt{(-4)^2+(3)^2}$

$YZ=\sqrt{16+9}$

$YZ=\sqrt{25}$

$YZ=5$

Length of XZ

Point X = $(x_1,y_1)=(-5,-1)$

Point Z = $(x_2,y_2)=(-9,-7)$

$XZ=\sqrt{(-9-(-5))^2+(-7-(-1))^2}$

$XZ=\sqrt{(-4)^2+(-6)^2}$

$XZ=\sqrt{16+36}$

$XZ=\sqrt{52}$

$XZ=7.2$

So, to find the area of triangle we will use heron's formula .

$Area =\sqrt{s(s-a)(s-b)(s-c)}$

Where $s=\frac{a+b+c}{2}$

a,b,c are sides of triangle

a=9

b=5

c=7.2

$s=\frac{9+5+7.2}{2}$

$s=10.6$

$Area =\sqrt{10.6(10.6-9-a)(10.6-5)(10.6-7.2)}$

$Area =\sqrt{322.9184}$

$Area =17.96$

Hence the area of triangle is 17.96 units squared