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Graph a line with the same slope as the line from problem 1# (hint the one i drew the line through already)And have it go through the point (3,-4) Identify the slope:And Y-intercept:Write the point slope equation of the line (y-y1=m(x-x1) ):

Graph a line with the same slope as the line from problem 1# (hint the one i drew-example-1
User Rajat Modi
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1 Answer

23 votes
23 votes

Firstly we need to find the slope of the plotted line;

we will pick two points on the line and apply the slope formula;


m=(y_2-y_1)/(x_2-x_1)

Let's select points;


(-9,1)\text{ and (9,7)}

Substituting into the formula, we have;


\begin{gathered} m=(y_2-y_1)/(x_2-x_1) \\ m=(7-1)/(9-(-9))=(6)/(9+9) \\ m=(6)/(18) \\ m=(1)/(3) \end{gathered}

Since we have the slope and a point (3,-4) on the new line we can use the point slope formula to find the equation of the line.


y-y_1=m(x-x_1)

Substituting the point and the slope,we have;


\begin{gathered} y-y_1=m(x-x_1) \\ y-(-4)=(1)/(3)(x-3) \\ y+4=(1)/(3)x-(3)/(3) \\ y+4=(1)/(3)x-1 \\ y=(1)/(3)x-1-4 \\ y=(1)/(3)x-5 \end{gathered}

From the equation, the y intercept is the point where x = 0.

y intercept is;


\begin{gathered} y=(1)/(3)x-5 \\ y=(1)/(3)(0)-5 \\ y=-5 \\ we\text{ have a point } \\ (0,-5) \end{gathered}

So we have two points on the new line.

Let us proceed to locate the two points and join to produce the new line;

Above is a graph of a line parallel to the line in problem 1, with the same slope and passes through point (3,-4).

The slope intercept equation of the line is;


y=(1)/(3)x-5

The point slope form of the equation is;


y+4=(1)/(3)(x-3)

The slope of the line is;


m=(1)/(3)

y-intercept is;


y=-5

Graph a line with the same slope as the line from problem 1# (hint the one i drew-example-1
User Dantuch
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3.4k points