Question

question
what is the equation of the line that passes through the point (-2, -1) and has a slope of \\(\frac{5}{2}\\)?

Explanation:

Step 1: Recall the point - slope form of a line

The point - slope form of a line is given by \(y - y_1=m(x - x_1)\), where \((x_1,y_1)\) is a point on the line and \(m\) is the slope of the line.

We are given that the point \((x_1,y_1)=(-2,-1)\) and the slope \(m = \frac{5}{2}\).

Step 2: Substitute the values into the point - slope form

Substitute \(x_1=-2\), \(y_1 = - 1\) and \(m=\frac{5}{2}\) into the formula \(y - y_1=m(x - x_1)\):

\(y-(-1)=\frac{5}{2}(x - (-2))\)

Simplify the left - hand side and the right - hand side:

The left - hand side \(y-(-1)=y + 1\)

The right - hand side \(\frac{5}{2}(x - (-2))=\frac{5}{2}(x + 2)\)

So we have \(y + 1=\frac{5}{2}(x + 2)\)

Step 3: Convert to slope - intercept form (optional, but common)

We can expand the right - hand side:

\(y+1=\frac{5}{2}x+\frac{5}{2}\times2\)

\(y + 1=\frac{5}{2}x+5\)

Subtract 1 from both sides:

\(y=\frac{5}{2}x+5 - 1\)

\(y=\frac{5}{2}x + 4\)

We can also leave it in the point - slope form or convert it to the standard form \(Ax+By = C\) (where \(A\), \(B\), and \(C\) are integers and \(A\geq0\)):

Starting from \(y + 1=\frac{5}{2}(x + 2)\)

Multiply both sides by 2 to get rid of the fraction:

\(2(y + 1)=5(x + 2)\)

Expand both sides:

\(2y+2 = 5x+10\)

Rearrange to get \(5x-2y=-8\)

Answer:

The equation of the line in point - slope form is \(y + 1=\frac{5}{2}(x + 2)\), in slope - intercept form is \(y=\frac{5}{2}x + 4\), and in standard form is \(5x-2y=-8\) (you can choose the form as per requirement, the most common is slope - intercept or point - slope. If we consider the slope - intercept form, the answer is \(y=\frac{5}{2}x + 4\))