Question

What is the equation of a line that passes though the points (0,85) and (4,8)?

Answer 1

Answer:

The equation of line is:

$77x+4y=340$

Step-by-step explanation:

Given points:

$(0,85)$ and $(4,8)$

To find the equation of the line.

Solution:

In order to find the equation of the line we will first find the slope of the line.

The slope of a line passing through points $(x_1,y_1)$ and $(x_2,y_2)$ the slope can be given as:

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

Plugging in the given points to find the slope of the line.

$m=\frac{8-85}{4-0}$

$m=\frac{-77}{4}$

Equation of line can be written in point slope form as:

$y-y_1=m(x-x_1)$

where $(x_1,y_1)$  is a point on the line.

Using point (4,8)

$y-8=-\frac{77}{4}(x-4)$

Multiplying both sides by 4.

$4(y-8)=4.(-\frac{77}{4})(x-4})$

$4(y-8)=-77(x-4)$

Using distribution:

$4y-32=-77x+308$

Subtracting $4y$ both sides.

$4y-4y-32=-77x-4y+308$

$-32=-77x-4y+308$

Subtracting 308 both sides.

$-32-308=-77x-4y+308-308$

$-340=-77x-4y$

Multiplying each term with -1.

$340=77x+4y$

Thus, equation of line is:

$77x+4y=340$