Equation of the line
[tex]\tt \Large \blue{\bold{y - y_1 = m (x - x_1)}}[/tex]
[tex]\:[/tex]
1. Find the equation of the line that passes through the point (-5,9) with slope of 5.
[tex]\tt \Large \rightarrow \blue{\bold{y - y_1 = m (x - x_1)}}[/tex]
[tex]\tt \large \rightarrow y - 9 = 5 (x - (-5))[/tex]
[tex]\tt \large \rightarrow y - 9 = 5 (x + 5)[/tex]
[tex]\tt \large \rightarrow y - 9 = 5x + 25[/tex]
[tex]\tt \large \rightarrow y = 5x + 25 + 9[/tex]
[tex]\tt \Large \rightarrow \purple{\bold{y= 5x + 34}}[/tex]
Answer:
[tex]\tt \Large \purple{\bold{y= 5x + 34}}[/tex]
[tex]\:[/tex]
2. Find the equation of the line that passes through the point (-7,5) with slope of ⅔.
[tex]\tt \Large \blue{\bold{y - y_1 = m (x - x_1)}}[/tex]
[tex]\tt \large \rightarrow y - 5 = \frac{2}{3} (x - (-7))[/tex]
[tex]\tt \large \rightarrow y - 5 = \frac{2}{3} (x + 7)[/tex]
[tex]\tt \large \rightarrow y - 5 = \frac{2}{3} \times x + \frac{2 \times 7}{3}[/tex]
[tex]\tt \large \rightarrow y - 5 = \frac{2}{3}x + \frac{14}{3}[/tex]
[tex]\tt \large \rightarrow y = \frac{2}{3}x + \frac{14}{3} + 5[/tex]
[tex]\tt \large \rightarrow y = \frac{2}{3}x + \frac{14}{3} + \frac{5 \times 3}{3}[/tex]
[tex]\tt \large \rightarrow y = \frac{2}{3}x + \frac{14}{3} + \frac{15}{3}[/tex]
[tex]\tt \Large \rightarrow \purple{\bold{y= \frac{2}{3}x + \frac{29}{3}}}[/tex]
Answer:
[tex]\tt \Large \purple{\bold{y= \frac{2}{3}x + \frac{29}{3}}}[/tex]
[tex]\:[/tex]
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