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Write the equation of the line in slope-intercept form. Through the points (-8, 15) and (6, 15)

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Question: Write the equation of the line in slope-intercept form. Through the points (-8, 15) and (6, 15)

Solution:

The equation of the line in slope-intercept form is :

y = mx + b

where m is the slope of the line, and b is the y-coordinate of the y-intercept of the line. Now, the slope of the line is given by the following formula:


m\text{ = }(Y2-Y1)/(X2-X1)

Where (X1,Y1) and (X2,Y2) are points on the line. In our case, we have that:

(X1, Y1) = (-8,15)

(X2,Y2) = (6,15)

Replacing these values in the equation of the slope we obtain:


m\text{ = }(Y2-Y1)/(X2-X1)\text{ =}(15-15)/(6-(-8))\text{ = 0}

then we have a horizontal line, because the slope is 0, for that, the equation of the line would be:

y = mx + b = 0(x) + b

then

y = b

now, take any point on the line, for example (x,y) = (6,15). Replacing this value in the previous equation, we obtain that the equation of the line is given by:


y\text{ = 15}

User Arjita Mitra
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