Find the equation of the line that passes through (1, 3) and (-1, 4).Find the equation of the line that passes through (1, 3) and (-1, 4).

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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We have the points (1,3)  and (-1,4)  passes through a line.

The equation for the line is:

y-y1 = m (x-x1)    where m is the slope.

m= (y2-y1)/(x2-x1)

    = (1/-2 = -1/2

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

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

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

y= (-1/2)x + 7/2

Multiply by 2:

2y = -x + 7

x + 2y - 7 = 0

tonys538's profile pic

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The equation of a line passing through two points `(x_1, y_1)` and `(x_2, y_2) ` is:

`(y - y_1)/(x - x_1) = (y_2 - y_1)/(x _2 - x_1)`

To determine the equation of a line passing through the points (1, 3) and (-1, 4) substitute `x_1 = 1` , `y_1 = 3` , `x_2 = -1` and `y_2 = 4` .

The equation of the line is:

`(y - 3)/(x - 1) = (4 - 3)/(-1 - 1)`

`(y - 3)/(x - 1) = -1/2`

2*(y - 3) = -1(x - 1)

2y - 6 = -x + 1

x + 2y - 7 = 0

The required equation of the line is x + 2y - 7 = 0

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll find the equation of the line that passes through the given points, using the formula:

(x2-x1)/(x-x1) = (y2-y1)/(y-y1)

Now, we'll substitute the coordinates of the given points, into the formula above:

(-1-1)/(x-1) = (4-3)/(y-3)

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

We'll cross multiply:

-2(y-3) = x-1

-2y + 6 = x-1

-2y = x-1-6

-2y = x-7

We'll divide by -2:

y = -x/2 + 7/2

So, the equation of the line is: y = -x/2 + 7/2

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

The line passing through the points (x1, y1) and (x2 y2) is given by:

(y-y1)/(y2-y1) = (x-x1)/(x2-x1).............(1)

(x1,y1) = (1,3) and (x2,y2) = (-1 , 4). Substituting these given coordinates in (1) we get:

(y-3)/(4-3) = (x-1)/(-1-1)

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

2(y-3) = (x-1)

2y-6 = x-1.

 x-1-2y+6 = 0

x-2y -5 = 0 is the line that passes through (1,3) and (-1,4).

krishna-agrawala's profile pic

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

A general equation of a line is:

y = mx + c

Where:

y = slope of the line

And c is a constant.

The slope of a line joining two points (x1, y1) and (x2, y2) is given by:

slope = m = (y2 - y1)/(x2 - x1)

Substituting given values of x1, y1, x2, and y2 in the above equation of slope:

m = (4 - 3)/( -1 - 1) = -1/2

Substituting this value of m in the general equation of line:

y = -x/2 + c   ... (1)

To find the value of c we substitute values of x1 and y1 in the above equation (1):

3 = -1/2 + c

==> c = 3 + 1/2 = 7/2

Substituting this value of c in equation (1):

y = -x/2 + 7/2

This equation can be simplified as:

x + 2y = 7

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