Question

Which equation is perpendicular to the line 3x+y=10 and passes through the point(6,-5)?

Answer 1

A perpendicular line's slope is the opposite of a parallel line's slope, so it would be 1/3.

1/3 times 6 is two, and subtract 7 to get -5.

Therefore, -7 is the y-intercept. 

So, the equation would be y=1/3x-7, so the answer is D.

Answer 2

Following are the solution to the line:

Given:

$\to line= 3x+y=10\\\\\to point= (6,-5)$

To find:

line=?

Solution:

$\to line= 3x+y=10\\\\\to point= (6,-5)$

line equation:

$\to y=mx+c$

Since the slope of a perpendicular line is indeed the inverse of a slope of a parallel line, it is $\frac{1}{3}$.

$\to \frac{1}{3} \times 6 =2,\ while\ 7 \neq-5.$

As just a result, the y-intercept is = $-7$. therefore, the "$y= \frac{1}{3}x-7$".

Therefore the final answer is "Option D"

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