Equation of line passing through $( 1,2 )$ and perpendicular to the line $2 x + 3 y = 1$.

i have attempted so far

to find a slope of line

$2x+3y=1$

$\left. \begin{array} { l } { \text { Slope m of a line of the form } A x + B y = C \text { equals } - \frac { A } { B } } \\ { A = 2 , B = 3 } \\ { m = - \frac { 2 } { 3 } } \end{array} \right.$

the point slope formula

$y - y _ { 1 } = m \left( x - x _ { 1 } \right)$

the point given (1,2)

$y - 2 = - \frac { 2 } { 3 } ( x - 1 )$

$y = - \frac { 2 } { 3 } x + \frac { 2 } { 3 } + \frac { 2 } { 1 } \times \frac { 3 } { 3 }$

$y = - \frac { 2 } { 3 } x + \frac { 2 } { 3 } + \frac { 6 } { 3 }$

$y = \frac { 2 } { 3 } x + \frac { 8 } { 3 }$

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 $-\frac23$. Therefore, the slope of the line that you're after is $\frac32\left(=-\frac1{-\frac23}\right)$. So, you're after a line $y=\frac32x+a$. Since you want it to pass through $(1,2)$, you want to have $2=\frac32+a$. So, take $a=\frac12$.

$$2x+3y=1\to y=\frac13-\frac23x$$ so the gradient of the line is $-\frac23$, which you got. The gradient of the perpendicular line is $$\frac{-1}{(-\frac23)}=\frac32$$

Thus the line is of the form: $$y=\frac32x+k$$ We know that $x=1\to y=2$, thus $$2=\frac32(1)+k\to k=?$$

• now i know where did i go wrong thank you very much Sep 5, 2018 at 17:38

Hint: The searched equation has the form $$y=\frac{3}{2}x+n$$. Plugging $x=1,y=2$ into this equation we get

$$2-\frac{3}{2}=n$$