Question

Find equation of line passing through the two points (3,5) and (-4,2)
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Answer 1

Answer:

$y=\frac{3}{7}x+3\frac{5}{7}$

Step-by-step explanation:

let (x_1,y_1)=(3,5) and (x_2,y_2)=(-4,2),

Then;

the slope,m, of the line connecting these two points is given as

$m=\frac{y_2-y_1}{x_2-x_1}\\ m=\frac{5-2}{3--4}\\ m=\frac{3}{7}$

To find the  equation of the line passing through these two points we use the following equation

$y-y_1=m(x-x_1)\\y-5=\frac{3}{7}*(x-3)\\ y=\frac{3}{7}x-\frac{9}{7}+5\\ y=\frac{3}{7}x+3\frac{5}{7}$

Therefore the equation of the line in the form y=mx+c is

$y=\frac{3}{7}x+3\frac{5}{7}$

Answer 2

The required equation is $3x-7y+26=0$

Step-by-step explanation:

Given : Line passing through the two points (3,5) and (-4,2).

To find : Equation of line ?

Solution :

The equation of line passing through two point is

$(y-y_1)=\frac{y_2-y_1}{x_2-x_1} (x-x_1)$

Here, $(x_1,y_1)=(3,5)$ and $(x_2,y_2)=(-4,2)$

Substitute in the equation,

$(y-5)=\frac{2-5}{-4-3} (x-3)$

$(y-5)=\frac{-3}{-7} (x-3)$

$(y-5)=\frac{3}{7} (x-3)$

$7y-35=3x-9$

$7y-3x=-9+35$

$7y-3x=26$

$3x-7y+26=0$

The required equation is $3x-7y+26=0$

#Learn more

Find equation of line passing through the two points (3,5) and (-4,2)

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