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L is the line $\\var{l[0]}x + \\var{l[1]}y = \\var{l[2]}$

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Find the point a, where L intersects the x-axis.

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a = ([[0]] , [[1]])

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Find the point b, where L intersects the y-axis.

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a = ([[0]] , [[1]])

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Find the equation of the line perpendicular to L and passing through c($\\var{px},\\var{py}$).

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y = [[0]]x + [[1]]

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Find the equation of the line parallel to L and passing through d($\\var{px1},\\var{py1}$).

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y = [[0]]x + [[1]]

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Find the mid-point of the line segment cd.

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Give answer to 2 decimal points.

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([[0]],[[1]])

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Calculate the distance from c to d.

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Give answer to 2 decimal places.

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ans = [[0]]

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$\\var{l[0]}x + \\var{l[1]}y = \\var{l[2]}$

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a) L intersects the x axis when $y=0$

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$\\var{l[0]}x + \\var{l[1]}(0) = \\var{l[2]}$

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$\\var{l[0]}x = \\var{l[2]}$

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$x = \\frac{\\var{l[2]}}{\\var{l[0]}}$

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Point $= (\\frac{\\var{l[2]}}{\\var{l[0]}},0)$

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b) L intersects the y axis when $x=0$

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$\\var{l[0]}0 + \\var{l[1]}y = \\var{l[2]}$

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$\\var{l[1]}y = \\var{l[2]}$

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$y = \\frac{\\var{l[2]}}{\\var{l[1]}}$

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Point $= (0,\\frac{\\var{l[2]}}{\\var{l[1]}})$

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c) To find the equation of a line we use the formula $y - y_1 = m(x - x_1)$, where $(x_1,y_1)$ is a point on the line and $m$ is the slope.

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We are given the point c($\\var{px},\\var{py}$). So, we need to find the slope $m$.

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Slope of L  = $-\\frac{\\var{l[0]}}{\\var{l[1]}}$. Since the line is perpendicular to L the product of the slopes is -1. i.e. $m\\times (-\\frac{\\var{l[0]}}{\\var{l[1]}})) = -1$

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$m = \\frac{\\var{l[1]}}{\\var{l[0]}}$

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$Y - (\\var{py}) = \\frac{\\var{l[1]}}{\\var{l[0]}}(X - (\\var{px}))$

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Rearranging we get

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$Y=\\frac{{\\var{l[1]}}}{{\\var{l[0]}}}x+\\frac{\\var{constant4a}}{\\var{l[0]}}$

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d) Using the slope of L from part c and the formula $y - y_1 = m(x - x_1)$, where $(x_1,y_1)$ is the point d$(\\var{px1},\\var{py1})$.

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Since the line is parallel to L the slope is the same.

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slope of L(m) = $-\\frac{\\var{l[0]}}{\\var{l[1]}}$

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$y - (\\var{py1}) = -\\frac{\\var{l[0]}}{\\var{l[1]}}$(x - \\var{px1})$\n$y= -\\frac{\\var{l[0]}}{\\var{l[1]}}x +  \\var{constanta}/\\var{l[1]}$\n e) Using the formula$(\\frac{x_1+x_2}{2},\\frac{y_1+y_2}{2})$, where$(x_1,y_1) = c(\\var{px},\\var{py})$and$(x_2,y_2) = d(\\var{px1},\\var{py1})$. \n$(\\frac{\\var{px}+\\var{px1}}{2},\\frac{\\var{py}+\\var{py1}}{2}) = (\\var{x5},\\var{y5})$\n f) Using the formula dis =$\\sqrt{(x_2-x_1)^2 + (y_2-y-1)^2}$, where$(x_1,y_1) = c(\\var{px},\\var{py})$and$(x_2,y_2) = d(\\var{px1},\\var{py1})$. \n$\\sqrt{(\\var{px1}-\\var{px})^2 + (\\var{py1})-(\\var{py})))^2}={\\var{dis}}\$

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Then round to 2 decimal places.

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Practice finding parallel and perpendicular lines to a given line.

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rebelmaths

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