Question

How do you find a standard form equation for the line with `(7,-4)` and perpendicular to the line whose equation is `x-7y-4=0`?

Answer 1

`7x+y=45`

Explanation:

`"the equation of a line in "color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))`
where A is a positive integer and B, C are integers.

`• " given a line with slope m then the slope of a line"`
`"perpendicular to it is"`

`•color(white)(x)m_(color(red)"perpendicular")=-1/m`

`"rearrange "x-7y-4=0" into "color(blue)"slope-intercept form"`

`•color(white)(x)y=mx+b`

`"where m is the slope and b the y-intercept"`

`rArry=1/7x-4/7rArrm=1/7`

`rArrm_(color(red)"perpendicular")=-1/(1/7)=-7`

`rArry=-7x+blarr" partial equation"`

`"to find b substitute "(7,-4)" into the partial equation"`

`-4=-49+brArrb=45`

`rArry=-7x+45larrcolor(red)" in slope-intercept form"`

`rArr7x+y=45larrcolor(red)" in standard form"`

Answer 2

`7x+y=45`

Explanation:

The given equation is in standard form which is `ax +by=c`

but we need to know its slope, so change it into the form `y=mx+c`

`x-7y-4 =0" "rarr x-4 =7y" "rarr7y = x-4`

`y=color(blue)(1/7)x -4/7`

`rarrm=1/7`

If lines are perpendicular then `m_1 xx m_2 = -1`

(One slope is the negative reciprocal of the other - flip and change the sign.)

The slope perpendicular to `1/7` is `-7`

Now we have a point `(7,-4)` and `m =-7` so substitute into the point-slope formula for a straight line.

`y-y_1 =m(x-x_1)`

`y-(-4) = -7(x-7)`

`y+4 = -7x+49" "larr` change to standard form:

`7x+y = 49-4`

`7x+y=45`