Respuesta :
Question 1
The equation of a line in point slope form with a slope of [tex]m= \frac{4}{3} [/tex] and passing through point [tex](7, \ 2)[/tex] is given by:
[tex]y-2= \frac{4}{3} (x-7)[/tex]
Question 2:
Given the line [tex]2x + 5y = 15\Rightarrow5y=-2x+15\Rightarrow y=- \frac{2}{5} x+3[/tex]
The equation of the line in point slope form passing through (4, -4) with a slope of [tex]m=- \frac{2}{5} [/tex] is given by:
[tex]y-(-4)=- \frac{2}{5} (x-4) \\ \\ \Rightarrow y+4=- \frac{2}{5} (x-4)[/tex]
Question 3:
Given the line [tex]-3x+y=8\Rightarrow y=3x+8\Rightarrow m =3[/tex].
The equation of the line in slope-intercept form passing through (-1, 5) with a slope of [tex]m=3[/tex] is given by:
[tex]y-5=3(x-(-1)) \\ \\ \Rightarrow y-5=3(x+1) \\ \\ \Rightarrow y-5=3x+3 \\ \\ \Rightarrow y=3x+8[/tex]
Question 4:
Given the line [tex]x-4y=4\Rightarrow 4y=x-4\Rightarrow y= \frac{1}{4} x-1 [/tex].
The equation of the line in slope-intercept form passing through (6, 3) with a slope of [tex]m=\frac{1}{4} [/tex] is given by:
[tex]y-3= \frac{1}{4} (x-6) \\ \\ \Rightarrow y=\frac{1}{4}x-\frac{6}{4}+3 \\ \\ \Rightarrow y=\frac{1}{4}x+\frac{3}{2}[/tex]
Question 5:
Given the line [tex]2x+3y=30\Rightarrow3y=-2x+30\Rightarrow y=- \frac{2}{3} x+10[/tex].
The equation of the line in standard form passing through (2, -5) with a slope of [tex]m=-\frac{2}{3} [/tex] is given by:
[tex]y-(-5)=- \frac{2}{3} (x-2) \\ \\ \Rightarrow3(y+5)=-2(x-2) \\ \\ \Rightarrow 3y+15=-2x+4 \\ \\ \Rightarrow 2x+3y=4-15 \\ \\ \Rightarrow2x+3y=-11[/tex]
Question 6:
Given the line [tex]y-5x+2=0\Rightarrow y=5x-2\Rightarrow m=5[/tex].
The equation of the line in standard form passing through (-3, -8) with a slope of [tex]m=5 [/tex] is given by:
[tex]y-(-8)=5(x-(-3)) \\ \\ \Rightarrow y+8=5x+15 \\ \\ \Rightarrow-5x+y=15-8 \\ \\ \Rightarrow-5x+y=7[/tex]
Question 7:
The line perpendicular to the line [tex]y=- \frac{1}{2} x+3[/tex] will have a slope of [tex]m=- \frac{1}{- \frac{1}{2} } =2[/tex].
The equation of a line in point slope form with a slope of [tex]m=2[/tex] and passing through point [tex](-4, \ 7)[/tex] is given by:
[tex]y-7=2(x-(-4)) \\ \Rightarrow y-7=2(x+4)[/tex]
Question 8:
Given the line [tex]8x-3y=12\Rightarrow3y=8x-12\Rightarrow y= \frac{8}{3} x-4[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{ \frac{8}{3} } =- \frac{3}{8} [/tex].
The equation of the line in point slope form passing through (6, -1) with a slope of [tex]m=- \frac{3}{8} [/tex] is given by:
[tex]y-(-1)=- \frac{3}{8} (x-6) \\ \\ \Rightarrow y+1=- \frac{3}{8} (x-6)[/tex]
Question 9:
Given the line [tex]2x+5y=10\Rightarrow5y=-2x+10\Rightarrow y=- \frac{2}{5} x+2[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{- \frac{2}{5} } = \frac{5}{2} [/tex].
The equation of the line in slope-intercept form passing through (4, 9) with a slope of [tex]m=\frac{5}{2}[/tex] is given by:
[tex]y-9= \frac{5}{2} (x-4) \\ \\ \Rightarrow y=\frac{5}{2} x-10+9 \\ \\ \Rightarrow y=\frac{5}{2} x-1[/tex]
Question 10:
Given the line [tex]6x-y-5=0\Rightarrow y=6x-5[/tex], the line perpendicular to the given line will have a slope of [tex]m=-\frac{1}{6} [/tex].
The equation of the line in slope-intercept form passing through (-3, 2) with a slope of [tex]m=-\frac{1}{6} [/tex] is given by:
[tex]y-2=- \frac{1}{6} (x-(-3)) \\ \\ \Rightarrow y=- \frac{1}{6} x- \frac{3}{6} +2 \\ \\ \Rightarrow y=- \frac{1}{6} x+ \frac{3}{2} [/tex]
Question 11:
Given the line [tex]4x+3y=24\Rightarrow3y=-4x+24\Rightarrow y=- \frac{4}{3} x+8[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{- \frac{4}{3} } = \frac{3}{4} [/tex].
The equation of the line in standard form passing through ((-5), 0) with a slope of [tex]m=\frac{3}{4} [/tex] is given by:
[tex]y-(-5)=- \frac{2}{3} (x-[tex]y-0= \frac{3}{4} (x-(-5)) \\ \\ \Rightarrow 4y=3(x+5)=3x+15 \\ \\ \Rightarrow-3x+4y=15[/tex]
Question 12:
Given the line [tex]2x-7y+21=0\Rightarrow7y=2x+21\Rightarrow \frac{2}{7} x+3[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{ \frac{2}{7} } =- \frac{7}{2} [/tex].
The equation of the line in standard form passing through (-1, -4) with a slope of [tex]m=-\frac{7}{2} [/tex] is given by:
[tex]y-(-4)=- \frac{7}{2} (x-(-1)) \\ \\ \Rightarrow y+4=- \frac{7}{2} (x+1) \\ \\ 2(y+4)=-7(x+1) \\ \\ 7x+2y=-7-8 \\ \\ \Rightarrow 7x+2y=-15[/tex]
Question 13:
Given the points (3, 2) and (9, 12), the equation of the line that passes throught the two points is given by:
[tex] \frac{y-2}{x-3} = \frac{12-2}{9-3} = \frac{10}{6} = \frac{5}{3} \\ \\ \Rightarrow3(y-2)=5(x-3) \\ \\ \Rightarrow3y-6=5x-15 \\ \\ \Rightarrow3y=5x-9 \\ \\ \Rightarrow y= \frac{5}{3} x-3[/tex]
The equation of a line in point slope form with a slope of [tex]m= \frac{4}{3} [/tex] and passing through point [tex](7, \ 2)[/tex] is given by:
[tex]y-2= \frac{4}{3} (x-7)[/tex]
Question 2:
Given the line [tex]2x + 5y = 15\Rightarrow5y=-2x+15\Rightarrow y=- \frac{2}{5} x+3[/tex]
The equation of the line in point slope form passing through (4, -4) with a slope of [tex]m=- \frac{2}{5} [/tex] is given by:
[tex]y-(-4)=- \frac{2}{5} (x-4) \\ \\ \Rightarrow y+4=- \frac{2}{5} (x-4)[/tex]
Question 3:
Given the line [tex]-3x+y=8\Rightarrow y=3x+8\Rightarrow m =3[/tex].
The equation of the line in slope-intercept form passing through (-1, 5) with a slope of [tex]m=3[/tex] is given by:
[tex]y-5=3(x-(-1)) \\ \\ \Rightarrow y-5=3(x+1) \\ \\ \Rightarrow y-5=3x+3 \\ \\ \Rightarrow y=3x+8[/tex]
Question 4:
Given the line [tex]x-4y=4\Rightarrow 4y=x-4\Rightarrow y= \frac{1}{4} x-1 [/tex].
The equation of the line in slope-intercept form passing through (6, 3) with a slope of [tex]m=\frac{1}{4} [/tex] is given by:
[tex]y-3= \frac{1}{4} (x-6) \\ \\ \Rightarrow y=\frac{1}{4}x-\frac{6}{4}+3 \\ \\ \Rightarrow y=\frac{1}{4}x+\frac{3}{2}[/tex]
Question 5:
Given the line [tex]2x+3y=30\Rightarrow3y=-2x+30\Rightarrow y=- \frac{2}{3} x+10[/tex].
The equation of the line in standard form passing through (2, -5) with a slope of [tex]m=-\frac{2}{3} [/tex] is given by:
[tex]y-(-5)=- \frac{2}{3} (x-2) \\ \\ \Rightarrow3(y+5)=-2(x-2) \\ \\ \Rightarrow 3y+15=-2x+4 \\ \\ \Rightarrow 2x+3y=4-15 \\ \\ \Rightarrow2x+3y=-11[/tex]
Question 6:
Given the line [tex]y-5x+2=0\Rightarrow y=5x-2\Rightarrow m=5[/tex].
The equation of the line in standard form passing through (-3, -8) with a slope of [tex]m=5 [/tex] is given by:
[tex]y-(-8)=5(x-(-3)) \\ \\ \Rightarrow y+8=5x+15 \\ \\ \Rightarrow-5x+y=15-8 \\ \\ \Rightarrow-5x+y=7[/tex]
Question 7:
The line perpendicular to the line [tex]y=- \frac{1}{2} x+3[/tex] will have a slope of [tex]m=- \frac{1}{- \frac{1}{2} } =2[/tex].
The equation of a line in point slope form with a slope of [tex]m=2[/tex] and passing through point [tex](-4, \ 7)[/tex] is given by:
[tex]y-7=2(x-(-4)) \\ \Rightarrow y-7=2(x+4)[/tex]
Question 8:
Given the line [tex]8x-3y=12\Rightarrow3y=8x-12\Rightarrow y= \frac{8}{3} x-4[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{ \frac{8}{3} } =- \frac{3}{8} [/tex].
The equation of the line in point slope form passing through (6, -1) with a slope of [tex]m=- \frac{3}{8} [/tex] is given by:
[tex]y-(-1)=- \frac{3}{8} (x-6) \\ \\ \Rightarrow y+1=- \frac{3}{8} (x-6)[/tex]
Question 9:
Given the line [tex]2x+5y=10\Rightarrow5y=-2x+10\Rightarrow y=- \frac{2}{5} x+2[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{- \frac{2}{5} } = \frac{5}{2} [/tex].
The equation of the line in slope-intercept form passing through (4, 9) with a slope of [tex]m=\frac{5}{2}[/tex] is given by:
[tex]y-9= \frac{5}{2} (x-4) \\ \\ \Rightarrow y=\frac{5}{2} x-10+9 \\ \\ \Rightarrow y=\frac{5}{2} x-1[/tex]
Question 10:
Given the line [tex]6x-y-5=0\Rightarrow y=6x-5[/tex], the line perpendicular to the given line will have a slope of [tex]m=-\frac{1}{6} [/tex].
The equation of the line in slope-intercept form passing through (-3, 2) with a slope of [tex]m=-\frac{1}{6} [/tex] is given by:
[tex]y-2=- \frac{1}{6} (x-(-3)) \\ \\ \Rightarrow y=- \frac{1}{6} x- \frac{3}{6} +2 \\ \\ \Rightarrow y=- \frac{1}{6} x+ \frac{3}{2} [/tex]
Question 11:
Given the line [tex]4x+3y=24\Rightarrow3y=-4x+24\Rightarrow y=- \frac{4}{3} x+8[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{- \frac{4}{3} } = \frac{3}{4} [/tex].
The equation of the line in standard form passing through ((-5), 0) with a slope of [tex]m=\frac{3}{4} [/tex] is given by:
[tex]y-(-5)=- \frac{2}{3} (x-[tex]y-0= \frac{3}{4} (x-(-5)) \\ \\ \Rightarrow 4y=3(x+5)=3x+15 \\ \\ \Rightarrow-3x+4y=15[/tex]
Question 12:
Given the line [tex]2x-7y+21=0\Rightarrow7y=2x+21\Rightarrow \frac{2}{7} x+3[/tex], the line perpendicular to the given line will have a slope of [tex]m=- \frac{1}{ \frac{2}{7} } =- \frac{7}{2} [/tex].
The equation of the line in standard form passing through (-1, -4) with a slope of [tex]m=-\frac{7}{2} [/tex] is given by:
[tex]y-(-4)=- \frac{7}{2} (x-(-1)) \\ \\ \Rightarrow y+4=- \frac{7}{2} (x+1) \\ \\ 2(y+4)=-7(x+1) \\ \\ 7x+2y=-7-8 \\ \\ \Rightarrow 7x+2y=-15[/tex]
Question 13:
Given the points (3, 2) and (9, 12), the equation of the line that passes throught the two points is given by:
[tex] \frac{y-2}{x-3} = \frac{12-2}{9-3} = \frac{10}{6} = \frac{5}{3} \\ \\ \Rightarrow3(y-2)=5(x-3) \\ \\ \Rightarrow3y-6=5x-15 \\ \\ \Rightarrow3y=5x-9 \\ \\ \Rightarrow y= \frac{5}{3} x-3[/tex]
