Question

What is an equation of the line that passes through the points (6, -5) and (3,0)?

Answer 1

Step-by-step explanation:

Hey there!

The equation of a st.line passing through points (6,-5) and (3,0) is;

$(y - y1) = \frac{y2 - y1}{x2 - x1} (x - x1)$

[Using double point formula]

Now, Put all values,

$(y + 5) = \frac{0 + 5}{3 - 6} (x - 6)$

Simplify it to get equation.

$(y + 5) = - \frac{5}{3} (x - 6)$

$3(y + 5) = - 5(x - 6)$

$3y + 15 = - 5x + 30$

$5x + 3y + 15 - 30 = 0$

$5x + 3y - 15 = 0$

Therefore the required equation is 5x + 3y - 15 = 0.

Hope it helps...

Answer 2

Answer:

y = -5/3x + 5

Step-by-step explanation:

The formula for the equation of a line is :

y - y1 = m (x-x1)

m is the gradient

finding the gradient

formula

$\frac{y2-y1}{x2-x1}$

= (0--5)/ (3-6)

=5 / -3

gradient = - 5/3

inserting the values in formula for the equation of a line

y - y1 = m (x-x1)

y - -5 = -5/3 (x - 6 )

y + 5 = -5/3 x + 10

y = -5/3x + 10 - 5

y = -5/3x + 5