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Answer :
We start with the equation of line [tex]\(PQ\)[/tex]:
[tex]$$\frac{3}{2} x + y - \frac{15}{2} = 0.$$[/tex]
Rearranging the equation to slope-intercept form, we obtain:
[tex]$$y = -\frac{3}{2}x + \frac{15}{2}.$$[/tex]
This shows that the slope of [tex]\(PQ\)[/tex] is
[tex]$$m_{PQ} = -\frac{3}{2}.$$[/tex]
Since line [tex]\(SR\)[/tex] is perpendicular to line [tex]\(PQ\)[/tex], its slope must be the negative reciprocal of [tex]\(m_{PQ}\)[/tex]. Thus, the slope of [tex]\(SR\)[/tex] is:
[tex]$$m_{SR} = \frac{2}{3}.$$[/tex]
The line [tex]\(SR\)[/tex] cuts the [tex]\(x\)[/tex]-axis at [tex]\(R\left(-\frac{3}{2}, 0\right)\)[/tex]. We can use this point to find the equation of [tex]\(SR\)[/tex] in the form
[tex]$$y = mx + c.$$[/tex]
Using the point-slope form:
[tex]$$y - y_1 = m(x - x_1),$$[/tex]
with [tex]\(m = \frac{2}{3}\)[/tex] and the point [tex]\(R\left(-\frac{3}{2}, 0\right)\)[/tex], we substitute:
[tex]$$y - 0 = \frac{2}{3}\left(x - \left(-\frac{3}{2}\right)\right) = \frac{2}{3}\left(x + \frac{3}{2}\right).$$[/tex]
Next, distribute the slope:
[tex]$$y = \frac{2}{3}x + \frac{2}{3} \cdot \frac{3}{2}.$$[/tex]
Simplify the constant term:
[tex]$$\frac{2}{3} \cdot \frac{3}{2} = 1.$$[/tex]
Thus, the equation becomes:
[tex]$$y = \frac{2}{3}x + 1.$$[/tex]
So, the values are:
- Slope [tex]\(m = \frac{2}{3}\)[/tex]
- [tex]\(y\)[/tex]-intercept [tex]\(c = 1\)[/tex]
Therefore, the equation of the line [tex]\(SR\)[/tex] is:
[tex]$$y = \frac{2}{3}x + 1.$$[/tex]
[tex]$$\frac{3}{2} x + y - \frac{15}{2} = 0.$$[/tex]
Rearranging the equation to slope-intercept form, we obtain:
[tex]$$y = -\frac{3}{2}x + \frac{15}{2}.$$[/tex]
This shows that the slope of [tex]\(PQ\)[/tex] is
[tex]$$m_{PQ} = -\frac{3}{2}.$$[/tex]
Since line [tex]\(SR\)[/tex] is perpendicular to line [tex]\(PQ\)[/tex], its slope must be the negative reciprocal of [tex]\(m_{PQ}\)[/tex]. Thus, the slope of [tex]\(SR\)[/tex] is:
[tex]$$m_{SR} = \frac{2}{3}.$$[/tex]
The line [tex]\(SR\)[/tex] cuts the [tex]\(x\)[/tex]-axis at [tex]\(R\left(-\frac{3}{2}, 0\right)\)[/tex]. We can use this point to find the equation of [tex]\(SR\)[/tex] in the form
[tex]$$y = mx + c.$$[/tex]
Using the point-slope form:
[tex]$$y - y_1 = m(x - x_1),$$[/tex]
with [tex]\(m = \frac{2}{3}\)[/tex] and the point [tex]\(R\left(-\frac{3}{2}, 0\right)\)[/tex], we substitute:
[tex]$$y - 0 = \frac{2}{3}\left(x - \left(-\frac{3}{2}\right)\right) = \frac{2}{3}\left(x + \frac{3}{2}\right).$$[/tex]
Next, distribute the slope:
[tex]$$y = \frac{2}{3}x + \frac{2}{3} \cdot \frac{3}{2}.$$[/tex]
Simplify the constant term:
[tex]$$\frac{2}{3} \cdot \frac{3}{2} = 1.$$[/tex]
Thus, the equation becomes:
[tex]$$y = \frac{2}{3}x + 1.$$[/tex]
So, the values are:
- Slope [tex]\(m = \frac{2}{3}\)[/tex]
- [tex]\(y\)[/tex]-intercept [tex]\(c = 1\)[/tex]
Therefore, the equation of the line [tex]\(SR\)[/tex] is:
[tex]$$y = \frac{2}{3}x + 1.$$[/tex]
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