Question

find the lines slope and a point on the line.
$y - 5 = -dfrac{3}{2}(x + 1)$
slope:
point on the line: $(quad,quad)$

Explanation:

Step1: Recall point - slope form

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

Step2: Identify slope and point from the given equation

The given equation is \(y - 5=-\frac{3}{2}(x + 1)\). We can rewrite \(x + 1\) as \(x-(- 1)\). So, comparing with the point - slope form \(y - y_1=m(x - x_1)\), we have:

• The slope \(m =-\frac{3}{2}\)
• The point \((x_1,y_1)=(-1,5)\) (since \(x_1=-1\) and \(y_1 = 5\))

Answer:

slope: \(-\frac{3}{2}\)
point on the line: \((-1,5)\)