Respuesta :
Lets do the first one(7). Fist change 10x -2y = 3,which is in standard form to point slope form so you can see the slope.
10x - 2y = 3
-2y = -10x + 3
-2y/-2 = -10x / -2 + 3/-2
y = 5x - 3/2
The slope is the m in y = mx + b so our slope is 5x, well it wants you to find the equation of a line that passes through (0,1) that is parallel to the equation this (Parallel) mean has the same slope.The slope is 5. Now that we know the slope is 5 we can find the equation by using the point slope form of an equation of a line.
point slope form = y – y1 = m(x – x1)
Remember m = slope = 5. Now plugin our x and y cord from the point (0,1) and our slope of 5
y - y1 = m(x - x1)
y - 1 = 5(x - 0)
y - 1 = 5x - 0
y - 1+1 = 5x + 1
y= 5x + 1
So the point (0,1) passes through y = 5x + 1
Number 8 is already in slope intercept form so you don't have to do anything to find the slope. You can see that the slope is 7 so just plug your point into y - y1 = m(x - x1) and then put it into slope intercept form.
8)
Point = (5,8)
Slope = 7
y - y1 = m(x - x1)
y - 8 = 7(x - 5)
y = 7x - 27
So for number 8 the point (5,8) passes through y = 7x - 27
(9) For nine they want an equation that is perpendicular, which means to flip the slope. For instance, if you have a slope of 2/5 the perpendicular slope would be 5/2. For 9 we see that the slope is 2/5 so its perpendicular slope would be 5/2. Now just plug your point into the point slope form equation.
Point = (-4, 6)
Slope = 2/5
perpendicular slope = 5/2
y - 6 = 5/2(x - (-4))
y = 5/2 x + 16
So point (-4, 6) passes through y = 5/2 x + 16
(10) Point (4,1) passes through y = 7/5 x - 23/5
first off, let's keep in mind something, two lines that are parallel to each other, have the same slope, and two lines that are perpendicular to each other, have negative reciprocal slopes.
7)
so is parallel to 10x-2y=3... now, let's solve that for "y", to put it in slope-intercept form
[tex]\bf 10x-2y=3\implies 10x-3=2y\implies \cfrac{10x-3}{2}=y \\\\\\ \stackrel{\textit{slope-intercept~form}}{\cfrac{5}{2}x-\cfrac{3}{2}}=y\impliedby \textit{slope is }\frac{5}{2}[/tex]
so then, we're looking for a line whose slope is 5/2 and passes (0,1)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 0}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=\cfrac{5}{2}(x-0)\implies y-1=\cfrac{5}{2}x \\\\\\ \boxed{y=\cfrac{5}{2}x+1}[/tex]
8)
now, is parallel to [tex]\bf y=\stackrel{slope}{7}x-6[/tex] , now, that's already in slope-intercept form, so the slope we can see is just 7, so then.
we are looking for a line whose slope is 7 passes (5,8)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 5}}\quad ,&{{ 8}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 7 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-8=7(x-5)\implies y-8=7x-35 \\\\\\ y=7x-35+8\implies \boxed{y=7x-27}[/tex]
9)
well, this function is also already in slope-intercept form, so we can tell the slope is 2/5.
now, the line we're looking for, is perpendicular to it, so it has a negative reciprocal slope, let's check
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{5}\\\\ slope=\cfrac{2}{{{ 5}}}\qquad negative\implies -\cfrac{2}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{2}[/tex]
so, we're looking for a line whose slope is -5/2 and passes (-4,6)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -4}}\quad ,&{{ 6}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-6=-\cfrac{5}{2}[x-(-4)] \\\\\\ y-6=-\cfrac{5}{2}(x+4)\implies y-6=-\cfrac{5}{2}x-10\implies \boxed{y=-\cfrac{5}{2}-4}[/tex]
10)
let's solve for "y" first
[tex]\bf 5x-7y=44\implies 5x-44=7y\implies \cfrac{5x-44}{7}=y \\\\\\ \stackrel{slope}{\cfrac{5}{7}}x-\cfrac{44}{7}=y[/tex]
now, let's find the negative reciprocal of that
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{5}{7}\\\\ slope=\cfrac{5}{{{ 7}}}\qquad negative\implies -\cfrac{5}{{{ 7}}}\qquad reciprocal\implies - \cfrac{{{ 7}}}{5}[/tex]
so, we're looking for a line whose slope is -7/5 and passes (4,1)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 4}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{7}{5} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=-\cfrac{7}{5}(x-4)\implies y-1=-\cfrac{7}{5}x+\cfrac{28}{5} \\\\\\ y=-\cfrac{7}{5}x+\cfrac{28}{5}+1\implies \boxed{y=-\cfrac{7}{5}x+\cfrac{33}{5}}[/tex]
7)
so is parallel to 10x-2y=3... now, let's solve that for "y", to put it in slope-intercept form
[tex]\bf 10x-2y=3\implies 10x-3=2y\implies \cfrac{10x-3}{2}=y \\\\\\ \stackrel{\textit{slope-intercept~form}}{\cfrac{5}{2}x-\cfrac{3}{2}}=y\impliedby \textit{slope is }\frac{5}{2}[/tex]
so then, we're looking for a line whose slope is 5/2 and passes (0,1)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 0}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies \cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=\cfrac{5}{2}(x-0)\implies y-1=\cfrac{5}{2}x \\\\\\ \boxed{y=\cfrac{5}{2}x+1}[/tex]
8)
now, is parallel to [tex]\bf y=\stackrel{slope}{7}x-6[/tex] , now, that's already in slope-intercept form, so the slope we can see is just 7, so then.
we are looking for a line whose slope is 7 passes (5,8)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 5}}\quad ,&{{ 8}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies 7 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-8=7(x-5)\implies y-8=7x-35 \\\\\\ y=7x-35+8\implies \boxed{y=7x-27}[/tex]
9)
well, this function is also already in slope-intercept form, so we can tell the slope is 2/5.
now, the line we're looking for, is perpendicular to it, so it has a negative reciprocal slope, let's check
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{2}{5}\\\\ slope=\cfrac{2}{{{ 5}}}\qquad negative\implies -\cfrac{2}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{2}[/tex]
so, we're looking for a line whose slope is -5/2 and passes (-4,6)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -4}}\quad ,&{{ 6}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{5}{2} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-6=-\cfrac{5}{2}[x-(-4)] \\\\\\ y-6=-\cfrac{5}{2}(x+4)\implies y-6=-\cfrac{5}{2}x-10\implies \boxed{y=-\cfrac{5}{2}-4}[/tex]
10)
let's solve for "y" first
[tex]\bf 5x-7y=44\implies 5x-44=7y\implies \cfrac{5x-44}{7}=y \\\\\\ \stackrel{slope}{\cfrac{5}{7}}x-\cfrac{44}{7}=y[/tex]
now, let's find the negative reciprocal of that
[tex]\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{5}{7}\\\\ slope=\cfrac{5}{{{ 7}}}\qquad negative\implies -\cfrac{5}{{{ 7}}}\qquad reciprocal\implies - \cfrac{{{ 7}}}{5}[/tex]
so, we're looking for a line whose slope is -7/5 and passes (4,1)
[tex]\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ 4}}\quad ,&{{ 1}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -\cfrac{7}{5} \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-1=-\cfrac{7}{5}(x-4)\implies y-1=-\cfrac{7}{5}x+\cfrac{28}{5} \\\\\\ y=-\cfrac{7}{5}x+\cfrac{28}{5}+1\implies \boxed{y=-\cfrac{7}{5}x+\cfrac{33}{5}}[/tex]
