Question 1024690
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Your answer to number 1 is sort of on the right track but your value for the constant term in the equation is off.


Your calculation of the slope of the perpendicular is correct, so I won't repeat that.  Knowing a point on a line and the slope, use the point-slope form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ y_1\ =\ m(x\ -\ x_1) ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 7\ =\ -\frac{4}{3}(x\ -\ 2)]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ -\ 7\ =\ -\frac{4}{3}x\ +\ \frac{8}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{4}{3}x\ +\ \frac{8}{3}\ +\ \frac{21}{3}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{4}{3}x\ +\ \frac{29}{3}]


2.  If point *[tex \Large Q] is on both lines, then since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ \frac{3}{4}x\ -\ 2]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ -\frac{4}{3}\ +\ \frac{29}{3}]


then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{4}x\ -\ 2\ =\ -\frac{4}{3}\ +\ \frac{29}{3}]


I'll let you convince yourself that *[tex \Large x\ =\ 5.6] and then *[tex \Large y\ =\ 2.2]


3.  Use the distance formula


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d\ =\ \sqrt{(x_1\ -\ x_2)^2\ +\ (y_1\ -\ y_2)^2}]


4.


*[illustration intersectingperpendicular.jpg]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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