Question

Find the equation of a straight line:
(i) with slope 2 and y-intercept 3;
(ii) with slope − 1/3 and y-intercept − 4.
(iii) with slope −2 and intersecting the x-axis at a distance of 3 units to the left of origin.

Answer 1

(i) Here, m = 2, c = 3

Substituting the values of m and c in y = mx + c, we get,

y = 2x + 3

Hence, the equation of the straight line with slope 2 and y-intercept 3 is y = 2x + 3

(ii) Here, $m=-\frac{1}{3},c=-4$

Substituting the values of m and c in y = mx + c, we get,

$$\begin{gathered} y=-\frac{x}{3}-4 \\ \Rightarrow x+3y+12=0 \end{gathered}$$

Hence, the equation of the straight line with slope $-\frac{1}{3}$ and y-intercept 4 is x + 3y + 12 = 0

(iii) Here, m = −2

Substituting the value of m in y = mx + c, we get,

y = −2x + c

It is given that the line y = −2x + c intersects the x-axis at a distance of 3 units to the left of the origin.
This means that the required line passes trough the point (−3, 0).

$$\begin{gathered} \therefore 0=-2\times \left(-3\right)+c \\ \Rightarrow c=-6 \end{gathered}$$

Hence, the equation of the required line is y = −2x − 6, i.e. 2x + y + 6 = 0