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3 years ago

Determine whether the systems have one solution, no solution, or infinitely many solutions

Mathematics

1 answer:

Lerok [7]3 years ago

5 0

Answer:

First system: no solution

Second system: one solution

Third system: one solution

Fourth system: infinite solutions

Fifth system: no solution

Step-by-step explanation:

First system: 3x-2y=3; 6x-4y=1

From the first equation: y = (3x - 3)/2

Using this value of y in the second equation:

6x - 6x + 6 = 1

6 = 1 -> System has no solution

Second system: 3x-5y=8; 5x-3y=2

From the first equation: x = (8 + 5y)/3

Using this value of x in the second equation:

5*(8 + 5y) - 9y = 6

40 + 25y - 9y = 6

16y = -34 -> y = -2.125

x = (8 - 5*2.125)/3 = -0.875

This system has one solution

Third system: 3x-2y=8; 4x-3y=1

From the first equation: x = (8 + 2y)/3

Using this value of x in the second equation:

4*(8 + 2y) - 9y = 3

32 + 8y - 9y = 6

y = 26

x = (8 + 2*26)/3 = 20

This system has one solution

Fourth system: 3x-6y=3; 2x-4y=2

From the first equation: x = 1 + 2y

Using this value of x in the second equation:

2*(1 + 2y) - 4y = 2

2 + 4y - 4y = 2

2 = 2

This system has infinite solutions

Fifth system: 3x-4y=2; 6x-8y=1

From the first equation: x = (2 + 4y)/3

Using this value of x in the second equation:

2*(2 + 4y) - 8y = 1

4 + 8y - 8y = 2

4 = 2

This system has no solution

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Triangle ABC has vertices at A(2,3),B(-4,-3) and C(2,-3) find the coordinates of each point of concurrency.

dem82 [27]

Answer:

Circumcenter =(-1,0)

Orthocenter =(2,-3)

Step-by-step explanation:

Given : Points A = (2,3), B = (-4,-3), C = (2,-3)

Formula used :

→Mid point of two points- $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

→Slope of two points - $\frac{y_2-y_1}{x_2-x_1})$

→Perpendicular of a line = $\frac{-1}{slope of line})$

Circumcenter- The point where the perpendicular bisectors of a triangle meets.

Orthocenter-The intersecting point for all the altitudes of the triangle.

To find out the circumcenter we have to solve any two bisector equations.

We solve for line AB and AC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = -1

Equation of AB with slope -1 and the coordinates (-1,0) is,

(y – 0) = -1(x – (-1))

y+x=-1………………(1)

Similarly, for AC

Mid point of AC = $(\frac{2+2}{2},\frac{3-3}{2})=(2,0)$

Slope of AC = $\frac{-3-3}{2-2}=\frac{-6}{0}$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of the perpendicular bisector = 0

Equation of AC with slope 0 and the coordinates (2,0) is,

(y – 0) = 0(x – 2)

y=0 ………………(2)

By solving equation (1) and (2),

put y=0 in equation (1)

y+x=-1

0+x=-1

⇒x=-1

So the circumcenter(P)= (-1,0)

To find the orthocenter we solve the intersections of altitudes.

We solve for line AB and BC

So, mid point of AB = $(\frac{2-4}{2},\frac{3-3}{2})=(-1,0)$

Slope of AB = $\frac{-3-3}{-4-2}=1$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of CF = -1

Equation of AB with slope -1 and the coordinates (-1,0) gives equation CF

(y – 0) = -1(x – (-1))

y+x=-1………………(3)

Similarly, mid point of BC = $(\frac{-4+2}{2},\frac{-3-3}{2})=(-1,-3)$

Slope of AB = $\frac{-3+3}{-4-2}=0$

Slope of the bisector is the negative reciprocal of the given slope.

So, the slope of AD = 0

Equation of AB with slope 0 and the coordinates (-1,-3) gives equation AD

(y-(-3)) = 0(x – (-1))

y+3=0

y=-3………………(4)

Solve equation (3) and (4),

Put y=-3 in equation (3)

y+x=-1

-3+x=-1

x=2

Therefore, orthocenter(O)= (2,-3)

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