# What is the area of the triangle formed by lines joining the points (-7, 0), (14, 0) and (12, 7)

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let the points be A(-7,0) ,B(14,0) and C(12,7)

we shall first find the lengths of the sides of the triangle ,

that is a = AB, b = BC, c = AC

let us use the coordinate system to find the lengths

AB = `sqrt((x2-x1)^2 +(y2-y1)^2)`

let (x1,y1) = (-7,0) and (x2,y2) = (14,0)

thus AB = `sqrt((14-(-7))^2+(0-0)^2)` = `sqrt((14+7)^2)` = `sqrt(21^2)` = 21

AB = 21 units

consider points B(14,0), C = (12,7)

similarly BC = `sqrt((12-14)^2 + (7-0)^2 )` = `sqrt((-2)^2 + 7^2)` = `sqrt(4+49)` =`sqrt(53)` = 7.28 units

A(-7,0) ,C(12,7)

and AC = `sqrt((12-(-7))^2+ (7-0)^2)` = `sqrt(19^2+7^2)` = `sqrt(361+49)` =`sqrt(410)` = 20.24 units

now area of the triangle = `sqrt(s(s-a)(s-b)(s-c))`

where s= (a+b+c) /2 = (21+7.28+20.24)/2 = 48.52/2= 24.26

s= 24.26

thus plug in a,b,c, and s to find the area

area = `sqrt(24.26(24.26-21)(24.26-7.28)(24.26-20.24))` = `sqrt(24.26*3.26*16.98*4.02)` = `sqrt(5389.57)` = 73.41square units

thus the area of the triangle = 73.41 square units

The area of the triangle formed by the lines joining the points (-7, 0), (14, 0) and (12, 7) has to be determined.

The points (-7, 0) and (14, 0) lie on the x-axis and the distance between them is 21. The point (12, 7) is at a perpendicular distance of 7 from the x-axis.

This gives the area of the triangle using the formula Area = (1/2)*base*height as (1/2)*21*7 = 73.5

**The area of the triangle formed by the lines joining the points (-7, 0), (14, 0) and (12, 7) is 73.5**