Answer :
Answer:
[tex]\textsf{a)}\;\;\; y = -\dfrac{5}{4}x+\dfrac{7}{2}[/tex]
[tex]\textsf{b)}\;\;\; y=\dfrac{4}{5}x+\dfrac{38}{5}[/tex]
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Step-by-step explanation:
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Part a
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If we consider line k as the transversal of lines m and n, then the same-side interior angles are 74° and 106°. Since these two angles add up to 180°, they are supplementary. According to the Same-Side Interior Angles Theorem, when a transversal crosses two lines and the same-side interior angles are supplementary, the two lines are parallel. Therefore, we can determine that lines m and n are parallel.
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Since lines m and n are parallel, they share the same slope. To find this slope, we can use the points (-3, -1) and (5, 2) from line m and substitute them into the slope formula:
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[tex]\text{Slope of line m}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{2-(-1)}{5-(-3)}=\dfrac{3}{8}[/tex]
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Therefore, the slope of line n is also 3/8.
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To find the equation of line n, substitute its slope, m = 3/8, and the point (4, 5) from line n into the point-slope form of a linear equation and rearrange into slope-intercept form:
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[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-5=\dfrac{3}{8}(x-4)[/tex]
[tex]y=\dfrac{3}{8}x+\dfrac{7}{2}[/tex]
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The slope-intercept form of a linear equation is y = mx + b, where m is the slope and b is the y-intercept. Therefore, the y-intercept of line n is (0, 7/2).
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Since line k intersects line n at its y-intercept, to find the slope of line k, substitute the y-intercept of line n, (0, 7/2), and the point (-2, 6) from line k into the slope formula:
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[tex]\text{Slope of line k}=\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{7}{2}-6}{0-(-2)}=\dfrac{-\frac{5}{2}}{2}=-\dfrac{5}{4}[/tex]
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Since the y-intercept of line k is the same as that of line n, the equation of line k in slope-intercept form is:
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[tex]\large\boxed{y = -\dfrac{5}{4}x+\dfrac{7}{2}}[/tex]
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Part b
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Since line j is perpendicular to line k, its slope is the negative reciprocal of line k's slope. Therefore:
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[tex]\text{Slope of line j}=\dfrac{4}{5}[/tex]
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To find the equation of line j, substitute its slope, m = 4/5, and the point (-2, 6) from line j into the point-slope form of a linear equation:
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[tex]y-y_1=m(x-x_1)[/tex]
[tex]y-6=\dfrac{4}{5}(x-(-2))[/tex]
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Simplify and rearrange into slope-intercept form:
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[tex]y-6=\dfrac{4}{5}(x+2)[/tex]
[tex]y-6=\dfrac{4}{5}x+\dfrac{8}{5}[/tex]
[tex]y=\dfrac{4}{5}x+\dfrac{38}{5}[/tex]
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Therefore, the equation of line j in slope-intercept form is:
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[tex]\large\boxed{y=\dfrac{4}{5}x+\dfrac{38}{5}}[/tex]
