How do you use the point on the line and the slope of the line to find three additional points through which the line passes: Point: (5, -6) Slope: m = 1?

Question

Nathan Axel · Accepted Answer

(0,-11), (4,-7) and (6,-5)

First, we need to use the formula to create a linear equation on which these points will lie. #(y-y_1)=m(x-x_1)# #m# would be the gradient of the graph, which we have; #m=1# #x_1 and y_1# are a point on the graph, which we are given; #x_1=5# and #y_1=-6# Consequently, a linear equation of #(y-(-6))=1(x-(5))# #(y+6)=(x-5)# #y=x-11# plot{y=x-11 [-17.41, 22.59, -15.04, 4.96]} We can now locate three more points that fall on this line by changing the values of x and y.

Isaiah Anthony · Answer

undefined To find three additional points through which the line passes using the given point (5, -6) and slope $m = 1$, you can use the slope-intercept form of the equation for a line, which is $y = mx + b$, where $m$ is the slope and $b$ is the y-intercept.

First, substitute the given point (5, -6) into the equation and solve for $b$:
$$ -6 = 1(5) + b $$
$$ -6 = 5 + b $$
$$ b = -6 - 5 $$
$$ b = -11 $$

Now that you have the y-intercept, the equation of the line becomes $y = x - 11$.

To find additional points, you can choose values of $x$ and plug them into the equation to solve for $y$.

For example:
- If $x = 6$, then $y = 6 - 11 = -5$, so the point (6, -5) is on the line.
- If $x = 7$, then $y = 7 - 11 = -4$, so the point (7, -4) is on the line.
- If $x = 8$, then $y = 8 - 11 = -3$, so the point (8, -3) is on the line.

Therefore, the three additional points through which the line passes are (6, -5), (7, -4), and (8, -3).