# Find the equation of the straight lines that pass through the followin

Find the equation of the straight lines that pass through the following sets of points:
a) $$\displaystyle{\left({2};\ {4}\right)};\ {\left({4};\ {7}\right)}$$
b) $$\displaystyle{\left({3};\ -{5}\right)};\ {\left(-{2};\ {2}\right)}$$
c) $$\displaystyle{\left({1};\ {3}\right)};\ {\left(-{3};\ {1}\right)}$$

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barcelodurazo0q

Step 1
Equation of line that passes through the points $$\displaystyle{\left({x}_{{{1}}},\ {y}_{{{1}}}\right)}\ \ {\left({x}_{{{2}}},\ {y}_{{{2}}}\right)}$$ is:
$$\displaystyle{y}-{y}_{{{1}}}={\frac{{{\left({y}_{{{2}}}-{y}_{{{1}}}\right)}}}{{{\left({x}_{{{2}}}-{x}_{{{1}}}\right)}}}}{\left({\left({x}-{x}_{{{1}}}\right)}\right.}$$
where,
slope $$\displaystyle{m}={\frac{{{\left({y}_{{{2}}}-{y}_{{{1}}}\right)}}}{{{\left({x}_{{{2}}}-{x}_{{{1}}}\right)}}}}$$
Step 2
a) Given points $$\displaystyle{\left({2},\ {4}\right)}\ \ {\left({4},\ {7}\right)}$$
now
$$\displaystyle{x}_{{{1}}}={2}$$
$$\displaystyle{y}_{{{1}}}={4}$$
$$\displaystyle{x}_{{{2}}}={4}$$
$$\displaystyle{y}_{{{2}}}={7}$$
Equation of a line:
$$\displaystyle{y}-{4}={\frac{{{\left({7}-{4}\right)}}}{{{\left({4}-{2}\right)}}}}{\left({\left({x}-{2}\right)}\right.}$$
$$\displaystyle{y}-{4}={\frac{{{3}}}{{{2}}}}{\left({x}-{2}\right)}$$
$$\displaystyle{y}={\frac{{{3}}}{{{2}}}}{x}-{\frac{{{6}}}{{{2}}}}+{4}$$
$$\displaystyle{y}={\frac{{{3}}}{{{2}}}}{x}+{1}$$
Step 3
Given points $$\displaystyle{\left({3},\ -{5}\right)}\ \ {\left(-{2},\ {2}\right)}$$
now
$$\displaystyle{x}_{{{1}}}={3}$$
$$\displaystyle{y}_{{{1}}}=-{5}$$
$$\displaystyle{x}_{{{2}}}=-{2}$$
$$\displaystyle{y}_{{{2}}}={2}$$
Equation of a line:
$$\displaystyle{y}+{5}={\frac{{{\left({2}+{5}\right)}}}{{{\left(-{2}-{3}\right)}}}}{\left({x}-{3}\right)}$$
$$\displaystyle{y}+{5}=-{\frac{{{7}}}{{{5}}}}{\left({x}-{3}\right)}$$
$$\displaystyle{y}=-{\frac{{{7}}}{{{5}}}}{x}+{\frac{{{21}}}{{{5}}}}-{5}$$
$$\displaystyle{y}=-{\frac{{{7}}}{{{5}}}}{x}-{\frac{{{4}}}{{{5}}}}$$
Step 4
Given points $$\displaystyle{\left({1},\ {3}\right)}\ {\left(-{3},\ {1}\right)}$$
now
$$\displaystyle{x}_{{{1}}}={1}$$
$$\displaystyle{y}_{{{1}}}={3}$$
$$\displaystyle{x}_{{{2}}}=-{3}$$
$$\displaystyle{y}_{{{2}}}={1}$$
Equation of a line:
$$\displaystyle{y}-{3}={\frac{{{\left({1}-{3}\right)}}}{{{\left(-{3}-{1}\right)}}}}{\left({x}-{1}\right)}$$
$$\displaystyle{y}-{3}={\frac{{{1}}}{{{2}}}}{\left({x}-{1}\right)}$$
$$\displaystyle{y}={\frac{{{1}}}{{{2}}}}{x}-{\frac{{{1}}}{{{2}}}}+{3}$$
$$\displaystyle{y}={\frac{{{1}}}{{{2}}}}{x}+{\frac{{{5}}}{{{2}}}}$$