The value for b in y=mx+b,directly from the table w/out drawing a graph or going any calculations ?

2 answers:

5 0

The value of b is the y intercept or when the line crosses the y axis
this only happens when the value of x=0
so just look for the ordered pair in the table with x=0 and y=b
so if you had a table and saw
(0,3)
(x,y)
so the point where x=0 is at point (0,3) y=3
therefor b=3

basically look for the matching y value for when x=0

quester [9]4 years ago

5 0

The b value in y=mx+b is how many units up or down the y-axis you go from the origin. For example, if y=mx+2, you would move 2 units up, or if it was y=mx-3 you would move 3 units down.

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What is the area of ABC?

Salsk061 [2.6K]

Answer:

Area of triangle ABC = 10

Option C is correct.

Step-by-step explanation:

We need to find area of triangle ABC

We are given A(2,1) B(4,7) and C(6,3)

The formula used to find $Area \ of \ triangle=\frac{1}{2}base\times height$ ABC is:

In the given triangle base= AC and height = BC

First we need to find the distance between A and C and B and C

The formula used to find distance between AC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=2, y_1=1, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-2)^2+(3-1)^2}\\Distance=\sqrt{(4)^2+(2)^2}\\Distance=\sqrt{16+4}\\Distance=\sqrt{20}$

find distance between BC $Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

We have $x_1=4, y_1=7, x_2=6, y_2=3$

Putting values and finding distance

$Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\Distance=\sqrt{(6-4)^2+(3-7)^2}\\Distance=\sqrt{(2)^2+(4)^2}\\Distance=\sqrt{4+16}\\Distance=\sqrt{20}$

So, We have height = $\sqrt{20}$ and base = $\sqrt{20}$

Finding area of triangle

$Area \ of \ triangle=\frac{1}{2}base\times height\\Area \ of \ triangle=\frac{1}{2}\sqrt{20} \times \sqrt{20} \\Area \ of \ triangle=\frac{1}{2}\times 20\\Area \ of \ triangle=10$