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stiv31 [10]
3 years ago
7

Which equation represents the line that passes through the points (–3, 7) and (9, –1)?

Mathematics
2 answers:
Sergeu [11.5K]3 years ago
4 0

The equation of the line that passes through the points \left({ - 3,7}\right) and \left({9, - 1}\right) is \boxed{2x + 3y = 15} .

Further explanation:

The linear equation with slope m and intercept c is given as follows.

\boxed{y = mx + c}

The formula for slope of line with points \left({{x_1},{y_1}}\right) and \left({{x_2},{y_2}}\right) can be expressed as,

\boxed{m=\frac{{{y_2}-{y_1}}}{{{x_2}-{x_1}}}}

Given:

The passes through the points are \left({ - 3,7}\right) and \left({9, - 1}\right) .

Explanation:

The line passes the points that are \left({ - 3,7}\right) and \left({9, - 1}\right).

The slope of the line can be obtained as follows.

\begin{aligned}m&=\frac{{ - 1 - 7}}{{9-\left({ - 3}\right)}}\\&=\frac{{ - 8}}{{12}}\\&=-\frac{2}{3}\\\end{aligned}

The slope of the line is m =  - \frac{2}{3}.

The line passes through the point \left({9, - 1}\right).

Substitute 9 for x, - \frac{2}{3} for m and -1 for y in equation y = mx + c to obtain the value of c.

\begin{aligned}-1&=\frac{{-2}}{3}\left(9\right)+c\\-1&=-2\left(3\right)+c\\-1&=-6+c\\-1+6&=c\\5&=c\\\end{aligned}

Hence, the equation of the line that passes through the points \left({ - 3,7}\right) and \left({9, - 1}\right) is \boxed{2x + 3y = 15}.

Learn more:

1. Learn more about inverse of the function <u>brainly.com/question/1632445. </u>

2. Learn more about equation of circle <u>brainly.com/question/1506955. </u>

3. Learn more about range and domain of the function <u>brainly.com/question/3412497 </u>

Answer details:

Grade: High School

Subject: Mathematics

Chapter: Linear equation

Keywords: numbers, slope, slope intercept, inequality, equation, linear inequality, shaded region, y-intercept, graph, representation, origin.

valina [46]3 years ago
3 0
First find the slope of the line.

m=slope=(y2-y1)/(x2-x1), in this case:

m=(-1-7)/(9--3)

m=-8/12

m=-2/3

You did not specify which form of the line you wanted.

Slope-intercept form is y=mx+b

y=-2x/3+b, using pint (9,-1) we can solve for b, the y-intercept...

-1=-2(9)/3+b

-1=-6+b

5=b  then we have:

y=-2x/3+5 or more neatly

y=(-2x+15)/3

Point-slope form of a line is:

y-y1=m(x-x1), again using (9,-1) as the point and m=-2/3 you get:

y+1=(-2/3)(x-9)

Standard form of a line is ax+by+c:

We can rearrange y=(-2x+15)/3 to standard form

y=(-2x+15)/3  multiply both sides by 3

3y=-2x+15  add 2x to both sides

2x+3y=15

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