Lines k and n intersect on the y-axis
a. What is the equation of line k?
(-2,6) 6-
(4,5)
b. What is the equation of line j?
710
m
(5.2)
2-
109
(-3,-1)
a. The equation of line k in slope-intercept form is
(Use integers or fractions for any numbers in the equation. Simplify your answer.)
Answers
a) The equation of line k is:
[tex]y = -\frac{202}{167}x + \frac{598}{167}[/tex]
b) The equation of line j is:
[tex]y = \frac{167}{202}x + \frac{1546}{202}[/tex]
The equation of a line, in slope-intercept formula, is given by:
[tex]y = mx + b[/tex]
In which:
m is the slope, which is the rate of change.b is the y-intercept, which is the value of y when x = 0.Item a:
Line k intersects line m with an angle of 109º, thus:[tex]\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}[/tex]
In which [tex]m_1[/tex] and [tex]m_2[/tex] are the slopes of k and m.
Line k goes through points (-3,-1) and (5,2), thus, it's slope is:[tex]m_1 = \frac{2 - (-1)}{5 - (-3)} = \frac{3}{8}[/tex]
The tangent of 109 degrees is [tex]\tan{109^{\circ}} = -\frac{29}{10}[/tex]Thus, the slope of line m is found solving the following equation:[tex]\tan{109^{\circ}} = \frac{m_2 - m_1}{1 + m_1m_2}[/tex]
[tex]-\frac{29}{10} = \frac{m_2 - \frac{3}{8}}{1 + \frac{3}{8}m_2}[/tex]
[tex]m_2 - \frac{3}{8} = -\frac{29}{10} - \frac{87}{80}m_2[/tex]
[tex]m_2 + \frac{87}{80}m_2 = -\frac{29}{10} + \frac{3}{8}[/tex]
[tex]\frac{167m_2}{80} = \frac{-202}{80}[/tex]
[tex]m_2 = -\frac{202}{167}[/tex]
Thus:
[tex]y = -\frac{202}{167}x + b[/tex]
It goes through point (-2,6), that is, when [tex]x = -2, y = 6[/tex], and this is used to find b.
[tex]y = -\frac{202}{167}x + b[/tex]
[tex]6 = -\frac{202}{167}(-2) + b[/tex]
[tex]b = 6 - \frac{404}{167}[/tex]
[tex]b = \frac{6(167)-404}{167}[/tex]
[tex]b = \frac{598}{167}[/tex]
Thus. the equation of line k, in slope-intercept formula, is:
[tex]y = -\frac{202}{167}x + \frac{598}{167}[/tex]
Item b:
Lines j and k intersect at an angle of 90º, thus they are perpendicular, which means that the multiplication of their slopes is -1.Thus, the slope of line j is:
[tex]-\frac{202}{167}m = -1[/tex]
[tex]m = \frac{167}{202}[/tex]
Then
[tex]y = \frac{167}{202}x + b[/tex]
Also goes through point (-2,6), thus:
[tex]6 = \frac{167}{202}(-2) + b[/tex]
[tex]b = \frac{(2)167 + 202(6)}{202}[/tex]
[tex]b = \frac{1546}{202}[/tex]
The equation of line j is:
[tex]y = \frac{167}{202}x + \frac{1546}{202}[/tex]
A similar problem is given at https://brainly.com/question/16302622
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