Question

Express y in terms of X in equation 2x-3y=12 find the points where the line represented by this equation cuts X axis and y axis​

Answer 1

We're given,

(i) $\longrightarrow \sf{2x - 3y = 12}$

$\longrightarrow \sf{2x - 12 = 3y}$

$\longrightarrow \sf{\dfrac{2x - 12}{3} = y}$

$\longrightarrow \sf{\dfrac{2x}{3} - \dfrac{12}{3} = y}$

$\longrightarrow \underline{\underline{\sf{y = \dfrac{2x}{3} - 4}}}$

$\sf{\\ \\}$

(ii) We have to find the point where the graph cuts x axis (x intercept) and the point where the graph cuts y axis (y intercept). For this, we have to use the Intercept Form of Straight Line

$\quad\quad\bullet \sf{\dfrac{x}{a} + \dfrac{y}{b} = 1}$

Here, a is the x intercept and b is the y Intercept.

So we have to express the given equation in this form.

$\longrightarrow \sf{2x - 3y = 12}$

$\longrightarrow \sf{\dfrac{2x - 3y}{12} = 1}$

$\longrightarrow \sf{\dfrac{2x}{12} - \dfrac{3y}{12} = 1 }$

$\longrightarrow \sf{\dfrac{x}{6} - \dfrac{y}{4} = 1}$

By comparing with $\sf{\dfrac{x}{a} + \dfrac{y}{b} = 1 }$, we can observe that,

$\longrightarrow \sf{a = 6}$

$\longrightarrow \sf{b = 4}$

Hence our answer is,

$\longrightarrow \underline{\underline{\sf{x\:intercept = 6}}}$

$\longrightarrow \underline{\underline{\sf{y\:intercept = 4}}}$

$\sf{\\}$

Extra:-

There are many forms of equations of straight line-

$\quad\quad\bullet\sf{ax + by = c \longrightarrow}$ Standard Equation of straight line

$\quad\quad\bullet\sf{y = mx + c \longrightarrow}$ Slope Intercept Form of straight line. Where m is the slope/gradient and c is the y intercept

$\quad\quad\bullet\sf{\dfrac{x}{a} + \dfrac{y}{b} = 1 \longrightarrow}$ Intercept Form of Straight Line. Where a is x intercept and b is y intercept

$\quad\quad\bullet\sf{y - {y}_{1} = m(x - {x}_{1}) \longrightarrow}$ Point slope form of straight line. Where m is the slope/gradient and $\sf{({x}_{1}, {y}_{1})}$ is a point on the graph.