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siniylev [52]
3 years ago
9

the midpoint of [KL] is (-8, 1). one endpoint is K(-6, 5). find the coordinates of the other endpoint L

Mathematics
1 answer:
ExtremeBDS [4]3 years ago
3 0

the correct question is

The midpoint of kl is m(–8, 1). one endpoint is k(–6, 5). find the coordinates of the other endpoint l.


we know that

the formula of midpoint is

Xm=(x1+x2)/2----> 2*Xm=x1+x2------> x2=2*Xm-x1

Ym=(y1+y2)/2----> 2*Ym=y1+y2------> y2=2*Ym-y1


let

(x1,y1)-------> (–6, 5).

(Xm,Ym)-----> (-8,1)


find (x2,y2)

x2=2*Xm-x1-----> 2*(-8)-(-6)----> -10

 y2=2*Ym-y1----> 2*(1)-5-----> -3


the point l is (-10,-3)




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without building the graph, find the coordinates of the point of intersection of the lines given by the equation y=3x-1 and 3x+y
DaniilM [7]
<h2><u>1. Determining the value of x and y:</u></h2>

Given equation(s):

  • y = 3x - 1
  • 3x + y = -7

To determine the point of intersection given by the two equations, it is required to know the x-value and the y-value of both equations. We can solve for the x and y variables through two methods.

<h3 /><h3><u>Method-1: Substitution method</u></h3>

Given value of the y-variable: 3x - 1

Substitute the given value of the y-variable into the second equation to determine the value of the x-variable.

\implies 3x + y = -7

\implies3x + (3x - 1) = -7

\implies3x + 3x - 1 = -7

Combine like terms as needed;

\implies 3x + 3x - 1 = -7

\implies 6x - 1 = -7

Add 1 to both sides of the equation;

\implies 6x - 1 + 1 = -7 + 1

\implies 6x = -6

Divide 6 to both sides of the equation;

\implies \dfrac{6x}{6}  = \dfrac{-6}{6}

\implies x = -1

Now, substitute the value of the x-variable into the expression that is equivalent to the y-variable.

\implies y = 3(-1) - 1

\implies     \ \ = -3 - 1

\implies     = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h3 /><h3><u>Method 2: System of equations</u></h3>

Convert the equations into slope intercept form;

\implies\left \{ {{y = 3x - 1} \atop {3x + y = -7}} \right.

\implies \left \{ {{y = 3x - 1} \atop {y = -3x - 7}} \right.

Clearly, we can see that "y" is isolated in both equations. Therefore, we can subtract the second equation from the first equation.

\implies \left \{ {{y = 3x - 1 } \atop {- (y = -3x - 7)}} \right.

\implies \left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

Now, we can cancel the "y-variable" as y - y is 0 and combine the equations into one equation by adding 3x to 3x and 7 to -1.

\implies\left \{ {{y = 3x - 1} \atop {-y = 3x + 7}} \right.

\implies 0 = (6x) + (6)

\implies0 = 6x + 6

This problem is now an algebraic problem. Isolate "x" to determine its value.

\implies 0 - 6 = 6x + 6 - 6

\implies -6 = 6x

\implies -1 = x

Like done in method 1, substitute the value of x into the first equation to determine the value of y.

\implies y = 3(-1) - 1

\implies y = -3 - 1

\implies y = -4

Therefore, the value(s) of the x-variable and the y-variable are;

\boxed{x = -1}   \boxed{y = -4}

<h2><u>2. Determining the intersection point;</u></h2>

The point on a coordinate plane is expressed as (x, y). Simply substitute the values of x and y to determine the intersection point given by the equations.

⇒ (x, y) ⇒ (-1, -4)

Therefore, the point of intersection is (-1, -4).

<h3>Graph:</h3>

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Need help presenting this solution on and number line and in interval notation. (I’m absolutely clueless haha)
beks73 [17]

Since the number is a fraction make a number line using fractions. The inequality sign includes being equal to, so add a solid dot on -5/2 and since x is less than that, draw an arrow pointing to the left.

See attached drawing.

For interval notation since x can be any number less than or equal to -5/2 it would be written as (-∞, -5/2]

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