find the equation of a line that contains this point (4, -3) and is perpendicular to the line -2x + 3y - 10= 0

Question

blacksteel · Answer

Let's start by putting the equation we need to be perpendicular to into y = mx + b form.
$$-2x + 3y - 10 = 0$$$$3y = 2x + 10$$$$y = \frac{ 2 }{ 3 }x + \frac{ 10 }{ 3 }$$

A straight line equation will always be perpendicular to any other straight line equation with the same slope. Therefore, we know that the equation we're writing will have a slope of 2/3. So our equation is$$y = \frac{ 2 }{ 3 }x + b$$ Now we plug in the provided x and y values from our point to solve for the b we need.$$-3 = \frac{ 2 }{ 3 }(4) + b =\frac{ 8 }{ 3 } + b$$$$b = -3  - \frac{ 8 }{ 3 } = \frac{ -9 }{ 3 } - \frac{ 8 }{ 3 } = \frac{ -17 }{ 3 }$$Plug this in and we get our line equation:$$y = \frac{ 2 }{ 3 }x - \frac{ 17 }{ 3 }$$