Question

)4x-3Y+12=0 intercept line and normal line​

Answer 1

★given:-

$\implies \: 4x - 3y + 12 = 0$

★find:-

• graph intercept line

★SOLUTION:-

$\implies \: 4x - 3y + 12 = 0 \\ \\ \implies 4x = 3y - 12\\ \\ \implies \star\pink{x = \frac{3y - 12}{4} }$

when y=4

$\implies \: x = \frac{3(4) - 12}{4} \\ \\ \implies \: x = \frac{12 - 12}{4} \\ \\ \implies \: x = \frac{0}{4} \\ \\ \implies \: x = 0$

A cordinet (x,y)=(0,4)

when y=0

$\implies \: x = \frac{3(0) - 12}{4} \\ \\ \implies \: x = \frac{ \cancel{-12}}{ \cancel{4}} \\ \\ \implies \: x = - 3$

B cordinet (x,y)=(-3,0)

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I hope it helps you