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3 years ago

Write the equation of a line that passes through (-2,6) and 1,-6)

2 answers:

6 0

Find slope first
(-6-6)/(1+2) = -12/3 = -4
Format: y = -4x + b
Plug in one point
6 = -4(-2) + b
6 = 8 + b, b = -2
Answer: y = -4x - 2

mote1985 [20]3 years ago

3 0

Answer: y=−2−4x

Check: Put 1,6 in the equation, -6 = -2-4(1). After solving, you get -6 (Check) So this equation is correct. Stay safe, and Happy Holidays!

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The line y =3x-5 meet x-axis at the point M. The line 3y+2x=2 meets y-axis at point N. Find the equation of the line joining M a

Law Incorporation [45]

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The line equation is,

$\boxed{\boxed{6x+15y-10=0}}$

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The line $y =3x-5$ meets x-axis at the point M, i.e M is the x-intercept of this line. At the x-intercept y=0, so

$\Rightarrow 0 =3x-5$

$\Rightarrow 3x=5$

$\Rightarrow x=\dfrac{5}{3}$

So, coordinate of M is $(\dfrac{5}{3},\ 0)$

The line $3y+2x=2$ meets y-axis at point N, i.e N is the y-intercept of this line. At the y-intercept x=0, so

$\Rightarrow 3y+2(0)=2$

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$\Rightarrow y=\dfrac{2}{3}$

So, coordinate of N is $(0,\ \dfrac{2}{3})$

The line joining M and N can be found out by applying two point formula of straight line,

$\Rightarrow \dfrac{y-y_1}{y_2-y_1}=\dfrac{x-x_1}{x_2-x_1}$

$\Rightarrow \dfrac{y-0}{\frac{2}{3}-0}=\dfrac{x-\frac{5}{3}}{0-\frac{5}{3}}$

$\Rightarrow \dfrac{y}{\frac{2}{3}}=\dfrac{x-\frac{5}{3}}{-\frac{5}{3}}$

$\Rightarrow -\dfrac{5}{3}y=\dfrac{2}{3}(x-\frac{5}{3})$

$\Rightarrow -5y=2(x-\dfrac{5}{3})$

$\Rightarrow -5y=2x-\dfrac{10}{3}$

$\Rightarrow 2x+5y-\dfrac{10}{3}=0$

As it is given that all the coefficients are integers, so multiplying with 3

$\Rightarrow 6x+15y-10=0$

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