Question

Write an equation of the line in the point slope form and y-intercept form. The line passes through (-4, 1) and Perpendicular to the line whose equation is x + 5y – 3 = 0

Answer 1

my name is Sindhu 5y=3-x

Answer 2

Step-by-step explanation:

Given :

To find the equation of the line, point slope form, y-intercept form, which passes through (-4,1) and perpendicular to the line whose equation is x + 5y - 3 = 0,.

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Let the equation of line be ax + by + c = 0,

Let a point with some co-ordinates on the line be (x,y)

From the given data,

The line passes through (-4,1)

For parallel lines slope is equal

$m_1 = \frac{y -y_{1} }{x - x_{1} } = \frac{y - (-4)}{x - 1} = \frac{y + 4}{ x - 1}$

We know that,

x + 5y - 3 = 0 (the equation of the line which is perpendicular to the given line.)

let x = 0.5 ,

x + 5y - 3 = 0

0.5 + 5y = 3 ⇒ 5y = 2.5

⇒ y = 0.5

∴ (0.5,0.5) is a solution of the equation of the line x + 5y - 3 = 0,

$m_2 = \frac{y - y_1}{x - x_1} = \frac{-4 - 0.5}{1 - 0.5}=  \frac{-4.5}{0.5} = -9$

We know that,

for perpendicular lines,

$m_1 m_2 = -1$

Then,

$(\frac{y+4}{x-1} ) (-9) = -1$

$9(\frac{y + 4}{x -1}) =  1$

$9y + 36 = x - 1$

$x - 9y - 37 = 0$

∴ x - 9y - 37 = 0 is the equation of the line.

point-slope form : $(y-y_1) = m(x-x_1)$

⇒$y - 0.5 = m(x - 0.5)$

y-intercept form :

When x = 0,

x - 9y - 37 = 0

0 - 9y - 37 = 0 ⇒ 9y = 37

The y-intercept is at y = $\frac{37}{9}$,. i.e., (0,$\frac{37}{9}$)