0 gradient of the line
3y + 2x - 5 = 0 is 3y = -2x + 5
y = -$\frac{2}{3}x$ + $\frac{5}{3}$
gradient of the other perpendicular line
through A(2,1) is $\frac{3}{2}$
therefore = $\frac{n-1}{4-2}$ = $\frac{3}{2}$
2n - 2 = 6;
2n = 8
n = 4
johnmulu answered the question on June 13, 2017 at 11:29
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