Question

(a) Check whether the line $3x-2y+9=0$ pass through the point $(1,6)$.
(b) Write down the equation of the line through $(3,7)$ and of slope $\frac{3}{2}$.
(c) Show that the lines mentioned above are parallel.

Answer 1

(a) The equation of the given line is 3x - 2y + 9 = 0
$\Rightarrow 3x-2y=-9$ ........(i)
To check whether (1, 6) lies on this line, we substitute x = 1 and y = 6 in equation (i).
L.H.S. $=3\left(1\right)-2\left(6\right)=3-12=-9$= R.H.S.
Thus, the point (1, 6) lies on the line 3x - 2y + 9 = 0, that is, the line 3x - 2y + 9 = 0 passes through the point (1, 6).
(b) (3, 7) is a point on a line whose slope is $\frac{3}{2}$
If (x, y) is another point on the same line, then, $\frac{y-7}{x-3}=\frac{3}{2}$
$3(x-3)=2(y-7)$
$\Rightarrow 3x-9=2y-14$
$\Rightarrow 3x-2y+5=0$
This is the required equation of the line.
(c) If $(x_1,y_1)$ is a point on the line 3x - 2y + 9 = 0, then
$3x_1-2y_1=-9$ ....... (i)
If $(x_2,y_2)$ is another point on that line, then
$3x_2-2y_2=-9$ ....... (ii)
From (i) and (ii),
$3x_1-2y_1=3x_2-2y_2$
$3x_2-3x_1=2y_2-2y_1$
$3(x_2-x_1)=2(y_2-y_1)$
Slop $=\frac{y_2-y_1}{x_2-x_1}=\frac{3}{2}$
Slope of the line $3x-2y+9=0$ is $\frac{3}{2}$
Since the slopes of the two lines are equal, the lines are parallel.