Question

Plot the lines:
(a)

3 x + 2 y = 0

(b)

x - 3 y + 6 = 0

Accepted Answer

The given equation of the line is $3 x + 2 y = 0$
$\Rightarrow 2 y = - 3 x$
$\Rightarrow y = - \frac{3 x}{2} \dots (1)$
Let us take two points to plot the curve,
if $x = 0$ , then $y = - \frac{3 (0)}{2} = 0$
So the point is $(0, 0)$
If $x = 2$ , then $y = - \frac{3 (2)}{2} = - 3$
So, the point is $(2, - 3)$
By denoting and joining these points in the graph.

$x - 3 y + 6 = 0 \Rightarrow 3 y = x + 6 ∴ y = \frac{x}{3} + 2$
Now, this equation is of the form $y = m x + c$ , where $m$ is the slope of the straight line and $c$ is the $y$ -intercept that is the $y$ -coordinate at which straight line cuts the $y$ -axis.
Here, $m = \frac{1}{3}, c = 2$ , since slope is positive the straight line makes an angle which is less than $90 °$ with the positive direction of $x$ -axis.
If $x = 0, y = 0 + 2 = 2 \Rightarrow (0, 2)$ is a point on the straight line.
If $x = 3, y = \frac{3}{3} + 2 = 1 + 2 = 3 \Rightarrow (3, 3)$ is a point on the straight line.
If $x = - 3, y = \frac{- 3}{3} + 2 = - 1 + 2 = 1 \Rightarrow (- 3, 1)$ is a point on the straight line.
Joining all these points, we get the required straight line.

Column I	Column II
i. Equation of side $A B$	a. $2 y + X = 2$
ii. Equation of side $E C$	b. $2 y - x = 2$
ill. Equation of side $C D$	c. $2 y + X = - 2$
iv. Equation of side $D A$	d. $2 y - x = - 2$

B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 17: CONCEPT APPLICATION EXERCISE 2.3

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 17: CONCEPT APPLICATION EXERCISE 2.3

Questions from B M Sharma Solutions for Chapter: Basic Mathematics, Exercise 17: CONCEPT APPLICATION EXERCISE 2.3 with Hints & Solutions

Learn and Prepare for any exam you want !