+2440 To figure out the equation of a line that passes through the given points (-6, -5) and (-4, -4), you must first know the standard form of a line. It is the following:
\(y=mx+b\)
Let m = slope of the line
Let b = the y-intercept (the point where the line touches the y-axis)
The first step is to figure out the slope of the line. How do we do that, you may ask? All you do is remember the slope formula.
\(m=\frac{y_2-y_1}{x_2-x_1}\)
We already have enough information to calculate the slope, m. We do this by substituting the given points into the formula.
\(m=\frac{-5-(-4)}{-6-(-4)}\)Simplify the fraction into simplest terms by evaluating the numerator and denominator separately.
\(m=\frac{-5+4}{-6+4}\)Of course, subtracting a negative is the same as adding a positive.
\(m=\frac{-1}{-2}\)The negatives in the numerator and denominator cancel each other out.
\(m=\frac{1}{2}\)
Great! We know the slope! Now, the only variable to figure out next is b, the y-intercept. We can do this by plugging in points of points on the line in the equation.
\(y=\frac{1}{2}x+b\)
In other words, to solve for b, you must plug in a point we know is one the line (either (-6,-5) or (-4,-4)) for x and y. I'll choose (-4,-4):
\(y=\frac{1}{2}x+b\)Plug in the coordinate (-4,-4) in its appropriate spots and then solve for b.
\(-4=\frac{1}{2}*-4+b\)Now, solve for b.
\(-4=-2+b\)Add 2 on both sides.
\(-2=b\)
Now that we have solved for both m and b, the equation that passes through the points (-6,-5) and (-4,-4) is \(y=\frac{1}{2}x-2\).
Do you need your answer in point-slope form? No problem! Remember the point-slope form
\(y-y_1=m(x-x_1)\)
Of course, m is the slope again. We have already calculated that. Let's substitute that in.
\(y-y_1=\frac{1}{2}(x-x_1)\)
\(y_1\hspace{1mm}\text{and}\hspace{1mm}x_1\) represent a point on the line. You can either substitute the first or the second set of coordinates. It doesn't matter. However, in the end, your answer should be one of these:
\(y+5=\frac{1}{2}(x+6)\)
\(y+4=\frac{1}{2}(x+4)\)
+2440 To figure out the equation of a line that passes through the given points (-6, -5) and (-4, -4), you must first know the standard form of a line. It is the following:
\(y=mx+b\)
Let m = slope of the line
Let b = the y-intercept (the point where the line touches the y-axis)
The first step is to figure out the slope of the line. How do we do that, you may ask? All you do is remember the slope formula.
\(m=\frac{y_2-y_1}{x_2-x_1}\)
We already have enough information to calculate the slope, m. We do this by substituting the given points into the formula.
\(m=\frac{-5-(-4)}{-6-(-4)}\)Simplify the fraction into simplest terms by evaluating the numerator and denominator separately.
\(m=\frac{-5+4}{-6+4}\)Of course, subtracting a negative is the same as adding a positive.
\(m=\frac{-1}{-2}\)The negatives in the numerator and denominator cancel each other out.
\(m=\frac{1}{2}\)
Great! We know the slope! Now, the only variable to figure out next is b, the y-intercept. We can do this by plugging in points of points on the line in the equation.
\(y=\frac{1}{2}x+b\)
In other words, to solve for b, you must plug in a point we know is one the line (either (-6,-5) or (-4,-4)) for x and y. I'll choose (-4,-4):
\(y=\frac{1}{2}x+b\)Plug in the coordinate (-4,-4) in its appropriate spots and then solve for b.
\(-4=\frac{1}{2}*-4+b\)Now, solve for b.
\(-4=-2+b\)Add 2 on both sides.
\(-2=b\)
Now that we have solved for both m and b, the equation that passes through the points (-6,-5) and (-4,-4) is \(y=\frac{1}{2}x-2\).
Do you need your answer in point-slope form? No problem! Remember the point-slope form
\(y-y_1=m(x-x_1)\)
Of course, m is the slope again. We have already calculated that. Let's substitute that in.
\(y-y_1=\frac{1}{2}(x-x_1)\)
\(y_1\hspace{1mm}\text{and}\hspace{1mm}x_1\) represent a point on the line. You can either substitute the first or the second set of coordinates. It doesn't matter. However, in the end, your answer should be one of these:
\(y+5=\frac{1}{2}(x+6)\)
\(y+4=\frac{1}{2}(x+4)\)
