What is the area of the triangle formed by lines y = 5-x, 2y

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).

Hope it's clear.

Thanks Bunnel for the detailed explanation! Very helpful..

We need 3 vertices. We can find them from the intersection of two lines.
y = 0 means that x can be anything . . . but set y = 0 and y= 5 - x equal.
The second equation "stops" y = 0 from going to the right infinitely.
Write equations for all three lines in slope-intercept form in order to keep intercepts straight.
y = mx + b
m = slope
b = y-intercept

Line 1
\(y = 5 - x\)
\(y = -x + 5\)

Line 2
\(y = 0\)
x can be any value. y is always 0
We need Line 1's x-intercept to give us a stop point

Line 3
\(2y = 3x\)
\(y = \frac{3}{2}x + 0\)

Any line that passes through the origin has slope-intercept form: \(y = mx\)
When x is 0, y is 0 and vice versa
The y-intercept is 0
The x-intercept is 0
So coordinates of one vertex are (0,0)

If that is not apparent, just find one of three vertices.
Line 2 = Line 3 will give you the same coordinates --
setting them equal means "this is the point at which the lines intersect"

Each vertex lies at the intersection point of TWO lines

To find that point, set equations equal, two at a time
We can do so because y = y

I think I see some confusion: "x" and "y" are generalized variables
we use, but the x's and y's are not all the same values.
I will use x and y to solve, but I will call
the exact counterparts of x,y coordinates
at the three different vertices (a,b), (c,d) and (e,f)

Vertex 1 (Line 1 equation = Line 3 equation):
\(-x + 5 = \frac{3}{2}x\)
\(-2x + 10 = 3x\)
\(10=5x\)
\(x=2\)
Plug THAT x-coordinate into EITHER equation to find y-coordinate of the intersection
\(y = (-x + 5) = (-2 + 5) = 3\)
Coordinates\((a,b)\) for that intersection point are \((2, 3)\)

Vertex 2 (Line 1 = Line 2) :
\(-x + 5 = 0\)
\(x = 5\)
Plug that x-value into \(y = -x + 5\)
Coordinates \((c,d)\) for that vertex are \((5,0)\)

Vertex 3 (Line 2 = Line 3):
\(\frac{3}{2}x = 0\)
\(x = 0\) (0 divided by anything except itself = 0)
Plug \(x = 0\) into \(y = \frac{3}{2}x\)
Coordinates (e,f) for that vertex are (0,0)

When two points have an x- or y-coordinate that is equal,
subtract one NON-equal coordinate from the other NON-equal coordinate.
That's length of one side of the triangle
(c,d): (5,0)
(e,f): (0,0)
The y-coordinate is 0. This side of the triangle lies on the x-axis.
\(x2 - x1 = (c - e)\) = length of a side
\((5 - 0) = 5\), which we can call the base of the triangle

The height of the triangle? Sketch if needed.
If the base lies on the x-axis, the height starts from . . . 0
Only one pair has a y-coordinate that is NOT 0
\((a,b) = (2, 3)\)
\(y2 - y1 = (b - 0) = (3 - 0) = 3\)
The triangle's height is \(3\)

Area =\(\frac{b*h}{2}\)
Area =\(\frac{5*3}{2}\)
Area = \(7.5\)

Answer A

dave13 -

Quote:

y = 5-x, so i plugged in this 2y = 3x [AND GOT X = 2] --> x = 2y/3 into this y = 5-x and got y = 3 so i have y= 3 and x =2

You found the intersection point for two lines.
You found coordinates for one vertex. Correct.
3 happens to be the height because the
base lies on y = 0, so the y-value, the height, is y2 - y1 = (3 - 0 ) = 3

But the base is not x=2.
No matter what y is, x always equals 2.
x = 2 is an infinite vertical line.
No side of this triangle is infinite.

What about the other two vertices?
What about the two other possible line intersections?
Plug y = 0 into y = 5 - x. Now what do you have?
It might prevent confusion if you write (x2,y2).
Hint: y2 = 0. When you plug y = 0 into that equation, you get a different x-value than the one you found above.

See my post; x and y are generalized variables but
each vertex has a specific different value of x,y
(I labeled each x,y set as a,b and c,d and e,f to clear confusion)
Plug y = 0 into 2y = 3x. Now what do you have?
It might help if you write (x3, y3)

Hope that helps!

There are many online graph plotters, try this one: https://graph-plotter.cours-de-math.eu/ or: https://www.wolframalpha.com/

Bunuel.. I got this question in gmat club tests..

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:
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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).

Answer: C.

No, we are equating to get the x-intercept of the point of intersection.

(2, 3) is the point of intersection of two lines: y = 5 - x and 2y = 3x.

You solve these two equations simultaneously.

y = 5 - x
Multiply by 2 to get: 2y = 10 - 2x .... (I)

2y = 3x ....(II)

Subtract (II) from (I)

0 = 10 - 5x
x = 2
Now substitute x = 2 in (I) to get y = 3
So x = 2 and y = 3 is the point where these two lines intersect.

I got the three co-ordinates as (5,5) (0,0) and (2,3)...
Anyone Please tel me where am i going wrong?

Bunuel, what do you use for making the graphs?

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

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)

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Kudos if the answer helped

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.

Merging similar topics. Please refer to the discussion above.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Pay attention to rules 1, 2, 3, 7 and 8. Thank you.

Hi Bunuel,

Please help me solidify the understanding for this - the reason you equated the line equations is because intersecting lines have the same slope right? Thank you.

hi Bunuel

i really dont get your solution so we have this y = 5-x, 2y = 3x, and y = 0 and we have these coordinaty geometry map with x and y axis

y = 5-x, so i plugged in this 2y = 3x --> x = 2y/3 into this y = 5-x and got y = 3 so i have y= 3 and x =2


hence y is height and x is basis ---> y = 3 and x = 2 i am sure you can visualize my wring solution

area (3*2)/2 = 3

pushpitkc - help pls:)

why are solving in a totally different way:? are quant gmatclub tests harder than the quant on gmat exam ?