Question

the equation of a straight line passing through (-3,-2) and perpendicular to the line y=x+3
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Answer 1

Given ,

The straight line passing through (-3,-2) and perpendicular to the line y=x+3

We know that ,

if equation of straight line is ax + by + c = 0 , then

$\boxed{ \tt{Slope \: (m) = - \frac{a}{b} }}$

if two lines are perpendicular to each other , then

$\boxed{ \tt{m_{1} \times m_{2} = - 1}}$

And the point slope form is given by

$\boxed{ \tt{Slope \: (m) = \frac{ y_{2} - y_{1}}{x_{2} - x_{1}} }}$

Thus , the slope of y = x + 3 or x - y + 3 = 0 will be

Slope (m) = -1/-1

Slope (m) = -1

Since , the straight line y = x + 3 is perpendicular to the line passing through (-3 , -2)

Thus ,

-1 × m' = -1

m' = 1

Now , the straight line passing through (-3,-2)

Thus ,

1 = (-2 - y)/(-3 - x)

-3 - x = -2 - y

x - y - 1 = 0

Therefore , the required equation of straight line is x - y - 1 = 0

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