Question

find the equation of the line which is perpendicular to the line 3x+2y-8=0 and passes through the midpoint of the line segment joining the points (6,4) and (4,-2)​

Answer 1

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Answer 2

2x - 3y - 7 = 0

Step-by-step explanation:

We are given the following in the question:

Equation of line:

$3x+2y-8=0\\2y = -3x + 8\\\\y = \dfrac{-3}{2}x + 4$

Comparing to general form of equation:

$y = mx + c$

We get,

$m = \dfrac{-3}{2}$

Thus, the slope of required line can be calculated as:

$m\times m_1 = -1\\\\\dfrac{-3}{2}\times m_1 = -1\\\\m_1 = \dfrac{2}{3}$

Midpoint of  (6,4) and (4,-2)​

$(x,y) = (\dfrac{6+4}{2}, \dfrac{4-2}{2}) = (5,1)$

Now, we know the slope and a point through which the line passes.

The equation of required line is:

$(y-y_1) = m_1(x -x_1)\\\\(y - 1) = \dfrac{2}{3}(x -5)\\\\3(y-1) = 2(x-5)\\3y-3 = 2x - 10\\2x-3y-7=0$

#LearnMore

Find the equation the line which passes through the midpoint of the line joining (2,-3) and (6,-7) and is parallel to 3x+4y+5=0​

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