Equation of line passing through (1,2) and perpendicular to the line 2x+3y=1. i have attempted so far to find a slope of line 2x+3y=1 Slope m of a line of the form Ax+By=C equals −AB, A=2,B=3 m=−23 the point slope formula y−y1=m(x−x1) the point given (1,2) y−2=−23x−1 y=−23x+23+21×33 y=−23x+23+63 y=23x+83 is this right answer ?

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Equation of line passing through (1,2) and perpendicular to the line 2x+3y=1.  i have attempted so far    to find a slope of line    2x+3y=1    Slope m of a line of the form Ax+By=C equals −AB, A=2,B=3  m=−23   the point slope formula    y−y1=m(x−x1)    the point given (1,2)    y−2=−23x−1    y=−23x+23+21×33    y=−23x+23+63    y=23x+83   is this right answer ?

Ian Adams · Accepted Answer

The line that you got is not orthogonal to 2x+3y=2. And it does not pass through (1,2).The slope of the line 2x+3y=2 is −23. Therefore, the slope of the line that you&#039;re after is 32(=−1−23). So, you&#039;re after a line y=32x+a. Since you want it to pass through (1,2), you want to have 2=32+a. So, take a=12

karton · Accepted Answer

2x+3y=1→y=13−23x so the gradient of the line is −23 , which you got. The gradient of the perpendicular line is −1(−23)=32 Thus the line is of the form: y=32x+k We know that x=1→y=2 ,thus 2=32(1)+k→k=?

Equation of line passing through (1,2) and perpendicular to the

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