What is the equation of a line that is perpendicular to −x+3y=9 and passes through the point (−3, 2) ?

1 answer:

7 0

• rearrange into form y = mx + c:
3y = 9 + x
y = 9/3 + x/3
y = 1/3x + 3

• substitute the gradient, 1/3, into the formula "m1 X m2 = -1" to find m2, which is the perpendicular of the gradient:
1/3 X m2 = -1
m2 = -3

• substitute x and y values of (-3, 2) into new equation, y = -3x + c, and rearrange, to find c and complete the equation:
2 = -3(-3) + c
2 = 9 + c
2 - 9 = c
c = -7

y = -3x - 7 is perpendicular to the line and passes through the point (-3, 2)

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<h2> $6 \sqrt{2} \: \: \: or \: \: \: 8.50 \: \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

$d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$

where

(x1 , y1) and (x2 , y2) are the points

From the question

The points are D(2, 0) and E(8, 6

The distance between them is

$|DE| = \sqrt{ ({2 - 8})^{2} + ({0 - 6})^{2} } \\ = \sqrt{( { - 6})^{2} + ( { - 6})^{2} } \\ = \sqrt{36 + 36} \\ = \sqrt{72} \: \: \: \: \: \: \: \: \: \: \\ = 6 \sqrt{2} \: \: \: \: \: \: \: \: \: \: \\ = 8.4852813$