Can I have help on the 10th question of this link?
Find slope-intercept form of the equation of the line perpendicular to 2x+3y=24 and passing throug (-6,2)
Standard Form of Equation of the line: y=mx+b
we have 2x+3y=24
y=(-2/3)x + 24/3
y=(-2/3)x + 8
so m = -2/3 and b = 8
Since lines are perpendicular multiplicatin of their slope will be (-1)
So slope of the required line will be 3/2
Now we have a point (-6,2) and slope 3/2 of the line we can easily find required lines by putting these values in the equation of the straight line poin-slope form.
y=mx+b we have y=2 , x=-6 and m=3/2 find b
2=(3/2)*(-6) + b
so slope-intercept form of the equation is y = (3/2)x + 11
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