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A line with a slope of -5 passes through the point 3 6 Find the area of the triangle in the first quadrant formed by this line and the coordinate axis Include step by step explanation of work?
Rosesare4ever09
First, let's find the equation of this line in the slope-intercept form, y = mx + b, in order to be able to graph this line.
By using the point-slope form we have:
(y - y1) = m(x - x1) using the point (3, 6) and slope, m = -5;
(y - 6) = -5(x - 3)
y - 6 = -5x + 15 add 6 to both sides;
y = -5x + 21
21 is the y- intercept, so we obtain another point, (0, 21)
By graphing the line with equation y = -5x + 21 we can find the zero of this function, which is the x-intercept. In our case x-intercept is 4.2 (you can use your calculator to find this x- intercept). Or you can make y equal to 0, and solve the equation 0 = -5x + 21 for x.
0 = -5x + 21 subtract 21 to both sides;
-21 = -5x divide by -5 to both sides;
4.2 = x
Now you can see that a right triangle is formed on the first quadrant, with length sides 21 and 4.2. Now you are able to find the area of this right triangle.
A = (1/2)(21)(4.2) = 44.1
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