Question

The equation of a straight line is
3x - 3y -7=0 find
you intercept of the line​

Answer 1

There are two forms of equation of straight line:-

$\quad \quad \bullet \sf{y = mx + c}$

where m is the slope/gradient of the graph and c is the y intercept (the point where the graph cuts the y axis). And,

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

where a is the x intercept and b is the y intercept

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We can solve this problem using both these equations.

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1st method:-

$\longrightarrow \sf{ 3x - 3y - 7 = 0}$

$\longrightarrow \sf{ 3y = 3x - 7 }$

$\longrightarrow \sf{ y = \dfrac{3x - 7}{3} }$

$\longrightarrow \sf{ y = x - \dfrac{7}{3} }$

$\longrightarrow \sf { y = 1x + \dfrac{-7}{3} }$

Comparing with $\sf{y = mx + c}$, we can observe that,

$\longrightarrow \sf{c = \dfrac{-7}{3} }$

And we know that c is the y intercept

$\longrightarrow \underline{\underline{\sf{y\:intercept = \dfrac{-7}{3}}}}$

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2nd method:-

$\longrightarrow \sf{ 3x - 3y - 7 = 0}$

$\longrightarrow \sf{3x - 3y = 7}$

$\longrightarrow \sf{\dfrac{3x - 3y}{7} = 1}$

$\longrightarrow \sf{ \dfrac{3x}{7} - \dfrac{3y}{7} = 1}$

$\longrightarrow \sf{ \dfrac{x}{(\dfrac{7}{3})} - \dfrac{y}{(\dfrac{7}{3})} = 1}$

$\longrightarrow \sf{ \dfrac{x}{(\dfrac{7}{3})} + \dfrac{y}{(\dfrac{-7}{3})} = 1}$

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

$\longrightarrow \sf{b = \dfrac{-7}{3} }$

And we know that b is the y intercept

$\longrightarrow \underline{\underline{\sf{y\:intercept = \dfrac{-7}{3}}}}$

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