# What is the area of the triangle formed by the lines y = 7 , x = 8 and 4x+ 3y =7?

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We are given the equations y=7, x= 8 and 4x + 3y = 7 which form the triangle. Now we know that y = 7 is a horizontal line and x = 8 is a vertical line. Now we need to find the lengths of the base and height and we can use the formula for area of a triangle Area = (1/2)* base * height.

We find where y=7 intersects 4x+ 3y =7 by substituting y with 7

=> 4x + 3*7 = 7

=> 4x + 21 = 7

=> 4x = -14

=> x = -14/4 = -7/2

So we can find the length of the base, it is 8+7/2 = 11.5

Now we need the point where x= 8 intersects 4x+ 3y =7.

Substituting

=> 4*8 + 3y =7

=> 32 + 3y = 7

=> 3y = 7-32

=> 3y = -25

=> y =-25/3

So now we can find the length of the height = 25/3+7 = 46/3

So the area is (1/2)* 11.5* 46/3 = 529/6

**Therefore the area of the triangle formed by the lines y=7, x= 8 and 4x + 3y = 7 is 529/6**

The triangle is formed by the lines y = 7 and x = 8 and 4x+3y = 7.

y = 7 and x = 8 intersect at ( 8, 7) at A say at right angles as both x= 8 and y = 7 are perpendicular to each other.

Y= 7 and 4x+3y = 7 intersect at the x coordinate given by:4x +3*7 = 7. Or 4x= 7- 21 = -14. So x = -14/4 = -3.5. So the point of intersection is (-3.5 , 7) ay at B.

4x+3y = 7 and x = 8 intersect at the y coordinate given by: 4*8+3y = 7. Or y = (7-32)/4 = -25/4 = -6.25. So the point of intersection is at C (8 ,-6.25)

Therefore the points A(8 , 7) , B(-3.5 , 7) and C(8 , -6.25) are the vertices of triangle.

AB = 8+3.5 = 11.5

AC = 7+6.25 = 13.25

BC^2 = (8+3.5)^2+(7+6.25)^2 which once again cofims that ABC is a right angled triangle with right angle at A, as AB^2+AC^2 = BC^2.

Therefore the area of the triangle = (1/2)AB*AC = (1/2)(11.5)(13.25) = 76.1875sq units.

To determine the area of the triangle formed intersecting the 3 given lines, we'll have to determine the length of the sides of the triangle.

We'll determine the vertices of the triangle first. The vertices of the triangles are the intercepting point of the given lines.

We'll determine the first vertex, namely the intercepting point of x=8 and 4x + 3y = 7.

We'll solve the system formed by the equations

x=8

4x + 3y = 7

We'll substitute x=8 in 4x + 3y = 7.

32 + 3y = 7

We'll subtract 32 both sides:

3y = 7 - 32

3y = 25

y = 25/3

So the first vertex is the point (8 , 25/3).

We'll determine the next vertex for y = 7 and 4x + 3y = 7.

We'll substitute y = 7 in 4x + 3y = 7.

4x + 21 = 7

We'll subtract 21:

4x = 7-21

4x = 14

x = 14/4

x = 7/2

The next vertex is (7/2 , 7)

The 3rd vertex is (8,7).

Since x is perpendicular to y, the triangle is right angled and the area of a right angled triangle is the half-product of the cathetus.

The base of triangle, AB, one cathetus, is located between (8,7) and (7/2 , 7).

The length of the base is:

AB = sqrt[(8-7.2)^2 + (7-7)^2]

**AB = 0.8**

The height of triangle, AB, one cathetus, is located between (8,7) and (8 , 25/3).

AC = sqrt [(8-8)^2+(25/3 - 7)^2]

**AC = 2/7**

The area of triangle ABC is:

A = AB*AC/2

A = (8*2)/2*10*7

A = 4/5*7

A = 4/35

**A = 114/1000 square units**