the sides of triangle are x=0 2x+3y-5 and 3x+2y+4 find the coordinates of the vertices of the triangle?

Answer 1

its a standard 3=4=5 triangle

Answer 2

Hi,
The first two vertices are easy to get as you have the first of the coordinates i.e. x =0

2x +3y - 5 =0
x=0
3y - 5 = 0
3y = 5
y = 5/3
First vertex: ( 0, 5/3)

Similarly for 3x + 2y = - 4
x=0
2y = -4
y = -2
Second vertex: (0,-2)

The third vertex is a little more difficult. It is the intersection of the two lines 2x+3y-5 = 0 and 3x + 2y + 4 +0.

2x + 3y - 5 =0
3x + 2y + 4=0
___________

Solve for x and y using the two simultaneous equations.
Multiply first line by 3 and the second line by 2. Subtract line 2 from line 1 and the x's cancel. Then left with only y values. Solve for y.

6x + 9y - 15 =0
6x + 4y + 8 = 0
____________
5y -23 = 0
5y = 23
y = 23/5
Now solve for x.
2x + 3y - 5 = 0
2x +3(23/5) - 5 = 0
Multipy across by 5
10x + 69 - 25 = 0
10 x = - 44
x = -22/5
Third vertex: (-22/5, 23/5)

So the vertices are: (0,5/3) (0,-2) and (-22/5, 23/5)


Oops, sorry.

Answer 3

The given sides should have been like
x=0 ..... (1)
2x+3y-5=0 .... (2)
3x+2y+4=0 .... (3) which are all in linear equations and solution is based on simultaneous equations

If we solve eqn no 1 and 2 we get
x=0 and y=5/3 that is (0,5/3)

if we solve eqn no 1 & 3 we get
x=0 and y=-2 that is (0,-2)

similarly if we solve eqn no 2 & 3 we get
x=-22/5 and y=23/5 that is (-22/5,23/5)

Therefore the vertices are (0,5/3) , (0.-2) & (-22/5,23/5)

Answer 4

Assuming that your equations are actually x=0, 2x+3y-5=0 and 3x+2y+4=0

Consider x=0 and 2x+3y-5=0, when x=0, 3y-5=0 => y=5/3
Consider x=0 and 3x+2y+4=0, when x=0, 2y+4=0 => y=-2
Consider 2x+3y-5=0 and 3x+2y+4=0, solve simultaneously.

2x+3y=5 -------(1)
3x+2y=-4 ------(2)

(1)x2, 4x+6y=10 --------(3)
(2)x3, 9x+6y=-12 -------(4)

(4)-(3), 5x = -22 => x = -22/5 = -4.4

Sub x = -4.4 into (1),
2(-4.4)+3y=5 => y = 4.6

So the coordinates of the vertices are (0, 5/3), (0, -2) and (-4.4, 4.6).

Answer 5

the equations of the given lines are

x=0......................(1)

2x+3y-5=0...........(2)

3x+2y+4=0..........(3)

line (1) intersects line (2) at the solution of equations (1) and(2)

sub x=0 into (2) we have 3y=5 >>>>> y=5/3 giving (0,5/3)

line (1) intersects line (3) at the solution of equations (1) and(3)

sub x=0 into (3) we have 2y= -4>>>>>y= -2 giving (0,-2)

line (2) intersects line (3) at the solution of equations (2) and(3)

6x+9y-15=0 mult (2) by 3

6x+4y+8 =0 mult (3) by 2

5y=23>>>>>>> y=23/5

6x + 92/5 +40/5 =0>>>>>6x= -132/5 >>>> x= -22/5

giving (-22/5,23/5)

since the points of intersection of the three lines are the vertices of the triangle,we have vertices at the points

(0,5/3),(0,-2) and(-22/5,23/5)

Answer 6

2(0)*3y-5=0
3y=5
y=5/3

3(0)+2y+4=0
2y=-4
y=-2

two coordinates are (0,5/3) and (0,-2)
I'm too lazy to find the third one

Answer 7

boring