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Answer:
2x + 3y -3=0
Step-by-step explanation:
The given equation of the line is ,
[tex]\implies 2x - 3y = 13 [/tex]
Now convert it into slope intercept form to get the slope , we get ,
[tex]\implies 3y = 2x - 13 \\\\\implies y =\dfrac{2}{3}x -\dfrac{13}{2}[/tex]
Therefore the slope is ,
[tex]\implies m = \dfrac{2}{3} [/tex]
We know that the product of slope of perpendicular lines is -1 . Therefore the slope of the perpendicular line will be ,
[tex]\implies m_{perpendicular}= -\dfrac{2}{3} [/tex]
Now one of the point is (-6,5) .On Using point slope form , we have ,
[tex]\implies y-y_1 = m( x - x_1) \\\\\implies y - 5 = -\dfrac{2}{3}( x + 6 ) \\\\\implies 3y - 15 = -2x -12
\\\\\implies 2x + 3y -15+12=0 \\\\\implies \underline{\underline{ 2x + 3y -3=0 }}[/tex]
Answer:
y = - [tex]\frac{3}{2}[/tex]x - 4
Step-by-step explanation:
2x – 3y = 13
3y = 2x + 13
y = [tex]\frac{2}{3}[/tex]x + [tex]\frac{13}{3}[/tex]
slope = 2/3
negative reciprocal = -3/2
y = -3/2x + b
(-6, 5)
5 = (-3/2)(-6) + b
5 = 9 + b
b = -4
y = -3/2x - 4