Hi, I'm really confused with perpendicular lines, how do I find the equation of the straight line? Please explain how to solve this. Thanks!
The equation of the straight line that passes through (0, -10) and perpendicular to the line with equation 3x-2y+5=0
The equation of the straight line that passes through (0, -10) and perpendicular to the line with equation 3x-2y+5=0
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First put the line in standard form:
y = (-3x-5)/(-2) ==> y = (3/2)x + 5/2.
Now we see that the slope is 3/2. If the second line is to be perpendicular to this one, it's slope moust be the negative reciprocal of the first. Therefore, the slope of the line we want is -2/3. Now since we know that it passes through the point (0,-1), we use the point-slope formula to get the equation of the line:
y - (-1) = (-2/3)(x - 0) ==> y = (-2/3)x - 1.
y = (-3x-5)/(-2) ==> y = (3/2)x + 5/2.
Now we see that the slope is 3/2. If the second line is to be perpendicular to this one, it's slope moust be the negative reciprocal of the first. Therefore, the slope of the line we want is -2/3. Now since we know that it passes through the point (0,-1), we use the point-slope formula to get the equation of the line:
y - (-1) = (-2/3)(x - 0) ==> y = (-2/3)x - 1.