Question

Write the equation of a line that is perpendicular to y=-0.3x+6 and that passes through the point (3,-8)

Answer 1

Hey there!

First, let's look at what perpendicular means. Imagine a cross, where there's all 90 degree angles. That's exactly what we're talking about when we say perpendicular. The given equation is in slope-intercept form, where we have:

y = mx + b
where m is the slope and b is the y-intercept.

When we're writing an equation with a perpendicular slope, we use the negative reciprocal of the given slope. Thus, we can make 0.3 1/3, and take the reciprocal to make 3, and make it negative 3 as it's the negative reciprocal. Now, we know we have a line with the slope of -3 and goes through (-3, 8). We can use the x and y values in this set of points, along with the slope, to create an equation to solve for b. That gives us:

8 = -3(-3) + b
8 = -9 + b
17 = b

Now, since we have slope and y-intercept, we can write our equation as:

y = -3x + 17

Hope this helps!

Answer 2

we know that

if two lines are perpendicular

then

the product of their slopes is equal to minus one

so

$m1*m2=-1\\m2=-1/m1$

in this problem the slope of the given line is

$m1=-0.3=-\frac{3}{10}$

the value of m2 is

$m2=-1/(-3/10)=\frac{10}{3}$

with the point $(3,-8)$ and the slope m2 find the equation of the line

the equation of the line in the form point-slope is

$y-y1=m*(x-x1)$

substitute the values

$y+8=\frac{10}{3}*(x-3)$

$y=\frac{10}{3}x-10-8$

$y=\frac{10}{3}x-18$

therefore

the answer is

$y=\frac{10}{3}x-18$