Question

Determine graphically the vertices of the triangle, the equations of whose sides are given below:
(i) 2y – x = 8, 5y – x = 14 and y – 2x = 1
(ii) y = x, y = 0 and 3x + 3y = 10

Answer 1

The graph is attached

(i) The vertices of the triangle is $(-4,2), (1,3), (2,5)$

(ii) The vertices of the triangle is $(0,0), (1.67,1.67), (3.33,0)$

Step-by-step explanation:

(i) The graph of the following lines is attached

$L1:2y-x=8$

$L2:5y-x=14$

$L3:y-2x=1$

From the graph it is clear that that the vertices of the triangle formed by the lines L1, L2 and L3 are $(-4,2), (1,3), (2,5)$

(ii) The graph of the following lines is attached

$L1:2y=x$

$L2:y=0$

$L3:3x+3y=10$

From the graph it is clear that that the vertices of the triangle formed by the lines L1, L2 and L3 are $(0,0), (1.67,1.67), (3.33,0)$

Know More:

Q: Find graphically the vertices of the triangle whose sides have the equation

2y–x=8, 5y–x=14, 2x–y= –1

Click Here: https://brainly.in/question/4140327

Q: Determine graphically the coordinates of the vertices of a triangle , the equation of whose sides are

y=x, 3y = x and x+y =8

Cick Here: https://brainly.in/question/2537768

Answer 2

(i)

Step-by-step explanation:

We have

2y−x=8

5y−x=14

y−2x=1

Now,

2y−x=8

⇒2y=8=x

⇒x=2y−8

When $$Y = 2$, we have

x=2×2−8=−4

When y=4, we have

x=2×2−8=−4

When y=4,wehavex=2×4−8=0

Thus, we have the following table giving points on the line 2y−x=8

x −4  0

y 2  4

Now,

5y−x=14

⇒5y−14=x

⇒x=5y−14

When y=2, we have

x=5×2−14=1

When y=3, we have

x=5×−14=1

Thus, we have the following table giving points on the line 5y−x=14

x  −4 1

y 2 3

We have

y−2x=1

⇒y−1=2x

⇒x=

2

y−1

​

When y=3, we have

x=

2

3−1

​

=1

When y=−1, We have

x=

2

−1−1

​

=1

Thus, we have the following table giving points on the line y−2x=1

x −1 1

y  1 3

Graph of the given equations:

Ref. image

From the graph of the lines represented by the given equations, we observe that the lines

Taken in pairs intersect each other at points A(−4,2),B(1,3) and C(2,5)

Hence, the vertices of the triangle are A(−4,2),B(1,3) and C(2,5),

for the given system of equations

solution