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Write the equation of a line perpendicular to y= 3/2x + 6 that passes through the point (-3,4)

2 Answers

2 votes

Answer: y = -2/3x + 2

Explanation:

When a line is perpendicular, it has the negative inverse gradient.

So,

Line 1: y = 3/2x + 6

Line 2: y = -2/3x + c

We can now substitute the values provided (-3,4),

Line 2: 4 = 6/3 + c

=> 4 = 2 + c

=> 2 = c

So,

Line 2: y = -2/3x + 2

User Masoud Keshavarz
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7.7k points
3 votes

Sure, let's go through this step by step.


1. Using y = mx + b, find the slope of the provided line. y = 3/2x + 6, means m is 3/2 in y = mx+ b, (the coefficient of x).

2. The slope of any line perpendicular to this line should be the negative reciprocal of the original slope. The reciprocal of 3/2 is 2/3, and then we take the negative of this to give the perpendicular slope. So, the slope of the line perpendicular to y = 3/2x + 6 is -2/3.

3. Now, we also know that our new line has to pass through the point (-3,4). We can use the point-slope form of the equation of the line, which is y - y1 = m(x - x1), where m is the slope (in this case, -2/3), and (x1, y1) is the point through which the line passes (in this case, (-3,4)).

4. Now, let's find the y-intercept (b in y = mx + b). Using the formula y = mx + b, b = y - mx and substituting slopes and coordinates of point, we'll get b = 4 - ((-2/3) * -3) = 2.

5. Thus, the equation of the line we are looking for is y = -2/3x + 2.

And that's your final answer! So, the equation of a line perpendicular to y = 3/2x + 6 and passing through the point (-3,4) is y = -2/3x + 2.

User Andrzej Polis
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7.8k points