Question

Find the equation of a line passing through (2, -1) and (3, -2). Write the equation in i) slope intercept form ii) intercepts form and iii) normal form. Also find the distance of the point (5, -3) to the line. ​

Answer 1

Dear student
$$\begin{gathered} \mathrm{Given}\mathrm{points}\mathrm{are}\mathrm{P}\left(2,-1\right)\mathrm{and}\mathrm{Q}\left(3,-2\right) \\ \mathrm{The}\mathrm{slope}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{is}\mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{-2+1}{3-2}=\frac{-1}{1}=-1 \\ \left[\because \mathrm{m}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}\right] \\ \mathrm{So}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{line}\mathrm{passing}\mathrm{through}\mathrm{P}\left(2,-1\right)\mathrm{and}\mathrm{Q}\left(3,-2\right)\mathrm{and}\mathrm{having}\mathrm{slope}\mathrm{m}=-1 \\ \mathrm{is} \\ \mathrm{y}-\left({\mathrm{y}}_1\right)=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right) \\ \Rightarrow \mathrm{y}-\left(-1\right)=-1\left(\mathrm{x}-2\right) \\ \Rightarrow \mathrm{y}+1=2-\mathrm{x} \\ \Rightarrow \mathrm{x}+\mathrm{y}=1 \\ \mathrm{Now}\mathrm{the}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{point}(5,-3)\mathrm{to}\mathrm{the}\mathrm{line}\mathrm{x}+\mathrm{y}-1=0\mathrm{is} \\ \mathrm{d}=\left|\frac{\mathrm{Ap}+\mathrm{Bq}+\mathrm{C}}{\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}}\right|,\mathrm{where}\mathrm{A}=1,\mathrm{B}=1,\mathrm{C}=-1\mathrm{and}\mathrm{p}=5\mathrm{and}\mathrm{q}=-3 \\ =\left|\frac{1\left(5\right)+1\left(-3\right)-1}{\sqrt{1+1}}\right| \\ =\left|\frac{5-3-1}{\sqrt{2}}\right| \\ =\frac{1}{\sqrt{2}}\mathrm{units} \end{gathered}$$
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