find the angle between the lines 3x +y-7=0 and x+2y+9=0
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slope of line ax + by + c = 0
=> m = -a/b
so,
slope of 3x + y - 7 = 0 is m1 = -3
& slope of x + 2y + 9 = 0 is m2 = -1/2
tanQ = | (m2 - m1) / (1 + m1m2) |
tanQ = | (-1/2 -3) / (1+(-1/2 * -3) |
tanQ = | (-7/2) / (1+ 3/2) |
tanQ = | (-7/2) / 5/2 |
tanQ = | -7 * 2 / 2 * 5 |
tanQ = | -7/5 |
tanQ = 7/5
So , Q = tan^(-1)[7/5]...
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=> m = -a/b
so,
slope of 3x + y - 7 = 0 is m1 = -3
& slope of x + 2y + 9 = 0 is m2 = -1/2
tanQ = | (m2 - m1) / (1 + m1m2) |
tanQ = | (-1/2 -3) / (1+(-1/2 * -3) |
tanQ = | (-7/2) / (1+ 3/2) |
tanQ = | (-7/2) / 5/2 |
tanQ = | -7 * 2 / 2 * 5 |
tanQ = | -7/5 |
tanQ = 7/5
So , Q = tan^(-1)[7/5]...
if Found Helpful Please Mark my answer Brainliest..
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