Question

Write in point-slope form an equation of the line that passes through the point (3,0) with slope $\frac{2}{3}$.
A.y=$\frac{2}{3}$x−2
B.y−3=$\frac{2}{3}$(x−0)
C.y−0=$\frac{2}{3}$(x−3)
D.y+0=$\frac{2}{3}$(x+3)

Answer 1

$\tt \: Equation \: of \: line \: passing \: through \: point \: (x_1 , y_1) \: and \: it's\: slope \: is \: m:$

$\boxed{\tt \large \longrightarrow y - y_1 = (x-x_1)m}$

Now , we have to find out the equation of the line that passes through the points (3, 0) and its slope is 2/3.

$\tt \large \implies y - 0 = (x-3)(\dfrac{2}{3})$

$\tt \large \implies 3y = 2(x-3)$

$\tt \large \implies 3y = 2x-6$

$\tt \large \implies 3y -2x+6=0$

Answer :

$\bf \large 3y -2x+6=0 \:or \:y - 0 = (x-3)(\dfrac{2}{3})$

Therefore , option (C) is correct.

$\tt \large \underline{\red{Additional-Information :}}$

$\tt \: Equation \: of \: line \: passing \: through \: points \: (x_1 , y_1) \: and \: (x_2 , y_2) :$

$\boxed{\tt \rightarrow y - y_1 = x-x_1\bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg )}$

$\boxed{\tt \longrightarrow \: here \: \bigg( \dfrac{y_2-y_1}{ x_2-x_1} \bigg ) is \: slope \: of \: line.}$