Question

Did you hear about the mathematician who wanted to make a fruit salad answers

Answer 1

Question 1

The equation of a line in point slope form with a slope of $m= \frac{4}{3}$ and passing through point $(7, \ 2)$ is given by:

$y-2= \frac{4}{3} (x-7)$



Question 2:

Given the line $2x + 5y = 15\Rightarrow5y=-2x+15\Rightarrow y=- \frac{2}{5} x+3$

The equation of the line in point slope form passing through (4, -4) with a slope of $m=- \frac{2}{5}$ is given by:

$y-(-4)=- \frac{2}{5} (x-4) \\ \\ \Rightarrow y+4=- \frac{2}{5} (x-4)$



Question 3:

Given the line $-3x+y=8\Rightarrow y=3x+8\Rightarrow m =3$.

The equation of the line in slope-intercept form passing through (-1, 5) with a slope of $m=3$ is given by:

$y-5=3(x-(-1)) \\ \\ \Rightarrow y-5=3(x+1) \\ \\ \Rightarrow y-5=3x+3 \\ \\ \Rightarrow y=3x+8$


Question 4:

Given the line $x-4y=4\Rightarrow 4y=x-4\Rightarrow y= \frac{1}{4} x-1$.

The equation of the line in slope-intercept form passing through (6, 3) with a slope of $m=\frac{1}{4}$ is given by:

$y-3= \frac{1}{4} (x-6) \\ \\ \Rightarrow y=\frac{1}{4}x-\frac{6}{4}+3 \\ \\ \Rightarrow y=\frac{1}{4}x+\frac{3}{2}$



Question 5:

Given the line $2x+3y=30\Rightarrow3y=-2x+30\Rightarrow y=- \frac{2}{3} x+10$.

The equation of the line in standard form passing through (2, -5) with a slope of $m=-\frac{2}{3}$ is given by:

$y-(-5)=- \frac{2}{3} (x-2) \\ \\ \Rightarrow3(y+5)=-2(x-2) \\ \\ \Rightarrow 3y+15=-2x+4 \\ \\ \Rightarrow 2x+3y=4-15 \\ \\ \Rightarrow2x+3y=-11$



Question 6:

Given the line $y-5x+2=0\Rightarrow y=5x-2\Rightarrow m=5$.

The equation of the line in standard form passing through (-3, -8) with a slope of $m=5$ is given by:

$y-(-8)=5(x-(-3)) \\ \\ \Rightarrow y+8=5x+15 \\ \\ \Rightarrow-5x+y=15-8 \\ \\ \Rightarrow-5x+y=7$



Question 7:

The line perpendicular to the line $y=- \frac{1}{2} x+3$ will have a slope of $m=- \frac{1}{- \frac{1}{2} } =2$.

The equation of a line in point slope form with a slope of $m=2$ and passing through point $(-4, \ 7)$ is given by:

$y-7=2(x-(-4)) \\ \Rightarrow y-7=2(x+4)$



Question 8:

Given the line $8x-3y=12\Rightarrow3y=8x-12\Rightarrow y= \frac{8}{3} x-4$, the line perpendicular to the given line will have a slope of $m=- \frac{1}{ \frac{8}{3} } =- \frac{3}{8}$.

The equation of the line in point slope form passing through (6, -1) with a slope of $m=- \frac{3}{8}$ is given by:

$y-(-1)=- \frac{3}{8} (x-6) \\ \\ \Rightarrow y+1=- \frac{3}{8} (x-6)$



Question 9:

Given the line $2x+5y=10\Rightarrow5y=-2x+10\Rightarrow y=- \frac{2}{5} x+2$, the line perpendicular to the given line will have a slope of $m=- \frac{1}{- \frac{2}{5} } = \frac{5}{2}$.

The equation of the line in slope-intercept form passing through (4, 9) with a slope of $m=\frac{5}{2}$ is given by:

$y-9= \frac{5}{2} (x-4) \\ \\ \Rightarrow y=\frac{5}{2} x-10+9 \\ \\ \Rightarrow y=\frac{5}{2} x-1$


Question 10:

Given the line $6x-y-5=0\Rightarrow y=6x-5$, the line perpendicular to the given line will have a slope of $m=-\frac{1}{6}$.

The equation of the line in slope-intercept form passing through (-3, 2) with a slope of $m=-\frac{1}{6}$ is given by:

$y-2=- \frac{1}{6} (x-(-3)) \\ \\ \Rightarrow y=- \frac{1}{6} x- \frac{3}{6} +2 \\ \\ \Rightarrow y=- \frac{1}{6} x+ \frac{3}{2}$



Question 11:

Given the line $4x+3y=24\Rightarrow3y=-4x+24\Rightarrow y=- \frac{4}{3} x+8$, the line perpendicular to the given line will have a slope of $m=- \frac{1}{- \frac{4}{3} } = \frac{3}{4}$.

The equation of the line in standard form passing through ((-5), 0) with a slope of $m=\frac{3}{4}$ is given by:

$y-(-5)=- \frac{2}{3} (x-[tex]y-0= \frac{3}{4} (x-(-5)) \\ \\ \Rightarrow 4y=3(x+5)=3x+15 \\ \\ \Rightarrow-3x+4y=15$



Question 12:

Given the line $2x-7y+21=0\Rightarrow7y=2x+21\Rightarrow \frac{2}{7} x+3$, the line perpendicular to the given line will have a slope of $m=- \frac{1}{ \frac{2}{7} } =- \frac{7}{2}$.

The equation of the line in standard form passing through (-1, -4) with a slope of $m=-\frac{7}{2}$ is given by:

$y-(-4)=- \frac{7}{2} (x-(-1)) \\ \\ \Rightarrow y+4=- \frac{7}{2} (x+1) \\ \\ 2(y+4)=-7(x+1) \\ \\ 7x+2y=-7-8 \\ \\ \Rightarrow 7x+2y=-15$



Question 13:

Given the points (3, 2) and (9, 12), the equation of the line that passes throught the two points is given by:

$\frac{y-2}{x-3} = \frac{12-2}{9-3} = \frac{10}{6} = \frac{5}{3} \\ \\ \Rightarrow3(y-2)=5(x-3) \\ \\ \Rightarrow3y-6=5x-15 \\ \\ \Rightarrow3y=5x-9 \\ \\ \Rightarrow y= \frac{5}{3} x-3$