Find an equation of the line satisfying the given conditions:
Through (6, 4); perpendicular to [tex]3X + 5Y = 38[/tex].
Answer (3)
To answer this, we will need to know:
• The slope of the equation we are trying to get • The point it passes through using the
First, we will need to find the slope of this equation. To find this, we must simplify the equation 3 x + 5 y = 38 into y = m x + b form. Lets do it!
3 x + 5 y = 38 = 5 y = − 3 x + 38 (Subtract 3x from both sides) = y = − 5 3 x + 5 38 (Divide both sides by 5)
The slope of a line perpendicular would have to multiply with the equation we just changed to equal -1. In other words, it would have to equal the negative reciprocal.
The negative reciprocal of the line given is 3 5 .
Now that we know the slope, we have to find out the rest of the equation using the slope formula, which is:
x − x 1 y − y 1 = m
Substituting values, we find that:
x − 6 y − 4 = 3 5
By simplifying this equation to slope-intercept form (By cross-multiplying then simplifying), we then get that:
y = 3 5 x − 6 , which is our final answer.
Thank you, and I wish you luck.
( 6 , 4 ) ; 3 x + 5 y = 38 s u b t r a c t 3 x f ro m e a c h s i d e 5 y = − 3 x + 8 d i v i d e e a c h \term b y 5 y = − 5 3 x + 5 38 T h e s l o p e i s : m 1 = − 5 3 I f m 1 an d m 2 a re t h e g r a d i e n t s o f tw o p er p e n d i c u l a r l in es w e ha v e m 1 ∗ m 2 = − 1
m 1 ⋅ m 2 = − 1 − 5 3 ⋅ m 2 = − 1 / ⋅ ( − 3 5 ) m 2 = 3 5
N o w yo u r e q u a t i o n o f l in e p a ss in g t h ro ug h ( 6 , 4 ) w o u l d b e : y = m 2 x + b 4 = 3 1 5 ⋅ 6 2 + b
4 = 5 ⋅ 2 + b 4 = 10 + b b = 4 − 10 b = − 6 y = 3 5 x − 6
The equation of the line that goes through the point (6, 4) and is perpendicular to the line given by 3 x + 5 y = 38 is y = 3 5 x − 6 . This was determined by first finding the slope of the original line and then using the point-slope form to formulate the new line's equation. By ensuring the slopes are negative reciprocals, we confirmed the lines are indeed perpendicular.
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