What is an equation of the line that passes through the points (-2, -5)and (-1, 1)?

Mathematics

2 answers:

damaskus [11]3 years ago

8 0

Answer: y = 6x + 7

Step-by-step explanation:

We can use slope formula to find the slope of a line that passes through two points (x₁ , y₁) and (x₂ , y₂)

Given points are: P(x₁ , y₁) = (-2, -5) and Q(x₂ , y₂) = (-1, 1)

Substituting these two known points in the slope formula, we have:

m = (1-(-5))/(-1-(-2)) = 1+5/-1+2 = 6/1 = 6

Now we can use the point-slope formula to write the equation of a line given a point on the line and the slope of the line:

m = 6 , given point P(x₁,y₁) = (-2, -5)

<h3>Formula = (y-y₁) = m(x-x₁) </h3>

(y-(-5)) = 6(x-(-2))

y+5 = 6x + 12

y = 6x + 12 -5

y = 6x + 7

Answer: y = 6x + 7

<h2><em>Spymore</em></h2>

suter [353]3 years ago

5 0

First find the slope(m)

Slope= y2-y1/x2-x1= (1-(-5))/(-1-(-2))= 6/1 = 6

Then find the y intercept (move up six and right one from (-1, 1) to get (0, 7)

The y-intercept is y=7

Now make the equation

Y= slope(x) + y-intercept

Y=6x+7

You might be interested in

Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

<span>The standard form of the equation of a circumference is given by the following expression:

</span> $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$
<span>
On the other hand, the general form is given as follows:

</span> $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ <span>

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

</span> $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

<em><u>First. a) matches 5) </u></em>

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

<em><u>Second. b) matches 1) </u></em>

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

<em><u>Third. c) matches 3)</u></em>

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

<em><u>Fourth. d) matches 6)</u></em>

6 0

3 years ago

Read 2 more answers

A train traveling at a constant speed covers 320 miles in 4 hours. What equation can be used to model the distance, d miles, the

vivado [14]

Answer:

80d:1t

Step-by-step explanation:

320mi:4h

80mi:1h

80d:1t

Every hour the train travels 80 miles