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The vertices of the triangle formed are (0,0), (5,0) and (2,3)

The first two vertices are fine, but how do we arrive at the third vertice of (2,3)? Since the equation for the third line is y = 0x + 0 / y = 0 it should be a horizontal line. Look forward to any explanations.

The vertices of the triangle formed are (0,0), (5,0) and (2,3)

The first two vertices are fine, but how do we arrive at the third vertice of (2,3)? Since the equation for the third line is y = 0x + 0 / y = 0 it should be a horizontal line. Look forward to any explanations.

We have three equations of lines: y_1=0 (x-axis), y_2=\frac{3x}{2}, y_3= 5-x.

Equate the functions to get the x-coordinate of the vertex (x-coordinate of the intersection point) and then substitute this value in either of functions to get y-coordinate of the vertex.

First vertex: y_1=0=y_2=\frac{3x}{2} --> x=0, y=0 --> vertex_1=(0,0); Second vertex: y_1=0=y_3=5-x --> x=5, y=0 --> vertex_2=(5,0); Third vertex: y_2=\frac{3x}{2}=y_3=5-x --> x=2, y=3 --> vertex_3=(2,3).

What is the area of a triangle enclosed by line 2x+3y=6, line y−x=2 and X-axis on the coordinate plane?

My question is.. why approach is different ?

If 2x+3y=6 and x=0 then y=2. And when y=0 then x=3.. (0,2) (3,0)

then y cant we do same with y=5-x? (5,0) (0,5)

and 2x=3y.. x=0 and y=0

where m wrong?

Below is that question. Please point out what differences are you talking about.

What is the area of a triangle enclosed by line 2x+3y=6, line y-x=2 and X-axis on the coordinate plane?

A. 3 B. 4 C. 5 D. 8 E. 10

Look at the diagram below:

Attachment:

Area.png [ 7.92 KiB | Viewed 997 times ]

Lines y=-\frac{2}{3}x+2 and y=x+2 intersect at point (0, 2). So the height of enclosed triangle is 2. Next, X-intercept of line y=-\frac{2}{3}x+2 is (3, 0) and X-intercept of line y=x+2 is (-2, 0), so the base of enclosed triangle is 3-(-2)=5. The area is \frac{1}{2}*base*height=\frac{1}{2}*5*2=5.

The vertices of the triangle formed are (0,0), (5,0) and (2,3)

The first two vertices are fine, but how do we arrive at the third vertice of (2,3)? Since the equation for the third line is y = 0x + 0 / y = 0 it should be a horizontal line. Look forward to any explanations.

We have three equations of lines: y_1=0 (x-axis), y_2=\frac{3x}{2}, y_3= 5-x.

Equate the functions to get the x-coordinate of the vertex (x-coordinate of the intersection point) and then substitute this value in either of functions to get y-coordinate of the vertex.

First vertex: y_1=0=y_2=\frac{3x}{2} --> x=0, y=0 --> vertex_1=(0,0); Second vertex: y_1=0=y_3=5-x --> x=5, y=0 --> vertex_2=(5,0); Third vertex: y_2=\frac{3x}{2}=y_3=5-x --> x=2, y=3 --> vertex_3=(2,3).

actually this highlighted thing is my question.. y have we taken another equation y=3/2x to get the points of equation y=5-x?

What is the area of a triangle enclosed by line 2x+3y=6, line y-x=2 and X-axis on the coordinate plane?

and i tuk this question (in bold text) to show u that , we didnt do same thing in this question. we tuk different vertices by individully solving an equation.

i tried my best to explain my question

thank u _________________

Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !

The vertices of the triangle formed are (0,0), (5,0) and (2,3)

The first two vertices are fine, but how do we arrive at the third vertice of (2,3)? Since the equation for the third line is y = 0x + 0 / y = 0 it should be a horizontal line. Look forward to any explanations.

We have three equations of lines: y_1=0 (x-axis), y_2=\frac{3x}{2}, y_3= 5-x.

Equate the functions to get the x-coordinate of the vertex (x-coordinate of the intersection point) and then substitute this value in either of functions to get y-coordinate of the vertex.

First vertex: y_1=0=y_2=\frac{3x}{2} --> x=0, y=0 --> vertex_1=(0,0); Second vertex: y_1=0=y_3=5-x --> x=5, y=0 --> vertex_2=(5,0); Third vertex: y_2=\frac{3x}{2}=y_3=5-x --> x=2, y=3 --> vertex_3=(2,3).

actually this highlighted thing is my question.. y have we taken another equation y=3/2x to get the points of equation y=5-x?

What is the area of a triangle enclosed by line 2x+3y=6, line y-x=2 and X-axis on the coordinate plane?

and i tuk this question (in bold text) to show u that , we didnt do same thing in this question. we tuk different vertices by individully solving an equation.

i tried my best to explain my question

thank u

The approaches are identical for both questions:

The base in both, is on x-axis. Find x-intercepts of other two lines and find the difference between them to get the length of the base. Next, find at what point do these two lines intersect. y-coordinate of that point would be the height.

To get intersection point of two lines y=3x/2 and y=5-x, we equate 3x/2 and 5-x to get x-coordinate of intersection point --> 3x/2=5-x --> x=2. Then we substitute it into either of the equations to get y=coordinate of intersection point --> y=5-2=3. So, y=3x/2 and y=5-x intersect at (2,3).

Re: Triangle problem - help me understand how to find vertices [#permalink]
12 Apr 2014, 12:03

rxn wrote:

Q: What is the area of the triangle formed by lines y=5−x, 2y=3x, and y=0?

Having trouble understanding the best approach to calculating vertices of the triangle which are given as follows:

(0,0) (5,0) (2,3)

Can anyone explain the best approach for calculating these? I'm fine with calculating the area once I have the vertices.

Thanks.

The vertices are the points where these lines intersect with each other. so, solve two two equation together:

point where y=0 and 2y = 3x intersect: put y=0 in 2y = 2x => x = 0, so coordinates (x,y) = (0,0) point where y = 0 and y = 5 - x intersect: put y=0 in y = 5-x => x = 5, so coordinates (x,y) = (5,0) point where 2y = 3x and y = 5 - x intersect: put y=3x/2 in y = 5-x => x = 2; y = 3, so coordinates (x,y) = (2,3)

-------------------------------- Kudos if the answer helped

Re: Triangle problem - help me understand how to find vertices [#permalink]
13 Apr 2014, 04:54

Alternatively, you could draw a quick graph on your scratch paper using the equations. For 2y= 3x , use the equation y= 3x/2 to draw the line. Then you just have to read the coordinates to have the coordinates of the vertices. I still think ind23's method is faster, but it is important to visualize geometry problems as it makes it easier to answer.

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