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helpmewithmath2012
I need help. I need to write an equation in slope intercept and standard form. Through (-5,-8) and perpendicular to y=1/2x+14
perpendicular means, has slope -2
An equation in slope intercept form is written as y=mx + c where 'm' is the slope and 'c' is the intercept. To write any equation in this form, you need the slope and one pair of points it passes through. Suppose the equation you are given represents line A and you have to find the equation of line B. If line A and libe B are perpendicular to each other, we have the following relation: \[slope of A = -1\div (Slope of B)\] Hence slope of line B will be -2 Next you use the following relation \[(y-y _{1}) = m(x-x_{1})\] Where y1 and x1 are the points you use. In this case, y1 is -8 and x1 is -5 Plus all these values in the relation above and you get y - (-8) = -2 (x -[-5]) => y +8 = -2(x+5) => y +8 = -2x -10 => y = -2x -18 where -18 is your slope
so how do i write it in standard form
y = -2x -18 This is standard form is often written as Ax + By = C Your standard form equation will be 2x + y = -18
Thank you so much!