Question

Which is the equation of a line that has a slope of -2/3 and passes through point (-3,-1)?

Answer 1

Answer:

Step-by-step explanation:

You have to use Point Slope Form:

• y - Y1 = m (x - X1)
• m is the slope
• Y1 & X1 is a point on the line
• The form allows you to identify the slope & the point on the line

About Problem:

• Since -2/3 is the slope, it represents m in y - Y1 = m (x - X1) form.

• -3 represents X1 in y - Y1 = m (x - X1) form

• -1 represents Y1  in y - Y1 = m (x - X1) form

y - Y1 = m (x - X1)

y - -1 = -2/3 (x - -3)  ---- This is in Point Slope Form

If you want to solve it, & put it in Slope Intercept form, it would look like this:

y = mx + b

y = -2/3 - 2 --- This is in Slope intercept Form.... I might've solved it wrong... I'm not sure...

Really really sorry if I'm incorrect...

Answer 2

Answer:  The required equation of the line is $2x+3y+9=0.$

Step-by-step explanation:  We are given to find the equation of a line that has a slope of $-\dfrac{2}{3}$ and passes through point (-3,-1).

SLOPE-INTERCEPT FORM :

The equation of a line having slope m and passing through the point (a, b) is given by

$y-b=m(x-a).$

For the given line, we have

$m=-\dfrac{2}{3},\\\\\\(a,b)=(-3,-1).$

Therefore, the equation of the line is given by

$y-b=m(x-a)\\\\\Rightarrow y-(-1)=-\dfrac{2}{3}(x-(-3))\\\\\\\Rightarrow y+1=-\dfrac{2}{3}(x+3)\\\\\Rightarrow 3y+3=-2x-6\\\\\Rightarrow 2x+3y+9=0.$

Thus, the required equation of the line is $2x+3y+9=0.$