Find the angle between the lines 3x+y+12=0 and x+2y-1=0 and the equations of the line drawn perpendicular to them from the point (3,-2)

Question

Vikas TU · Answer

Dear student slope of line = -A/B m1 = -3 m2 = -0.5 TanQ = m1m2/1-m1m2 Put the value of m1 and m2 Q = tan inverse (-3) Sope oof line perpendicular to line 1 = -1/3 Slope of line perpendicular to line 2 = -2 Equation 1 = y+2 = -1/3 (x-3) Equation 2 = y+2 = -2(x-3)

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Find the angle between the lines 3x+y+12=0 and x+2y-1=0 and the equations of the line drawn perpendicular to them from the point (3,-2)

Find the angle between the lines 3x+y+12=0 and x+2y-1=0 and the equations of the line drawn perpendicular to them from the point (3,-2)

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Find the angle between the lines 3x+y+12=0 and x+2y-1=0 and the equations of the line drawn perpendicular to them from the point (3,-2)

Find the angle between the lines 3x+y+12=0 and x+2y-1=0 and the equations of the line drawn perpendicular to them from the point (3,-2)

1 Answers

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Other Related Questions on 10 grade maths

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