+501 Hey guys I need help quick, !(-1,n), B(3,5),C(-4,0)andD(0,2) are 4 given points. Find the value of n in each of the following situations: AB//CD AB is perpendicular with CD P(a,1),Q(0,-a,R(2,1)andS(-1,-1) are 4 vertices of a quadrilateral. If PQ // SR, find the value of a, how many pairs of parallel sides does the quadrilateral PQRS have?
+2446 Here are the points that are given, \(A(-1,n)\hspace{3mm},B(3,5)\hspace{3mm},C(-4,0)\hspace{3mm},D(0,2)\). Our first task is to fid the value for n wherein \(\overline{AB}\hspace{1mm}||\hspace{1mm}\overline{CD}\). Let's think about what we must do first.
How can we verify if two segments are parallel to each other? We can compare the slopes! If the slopes are the same, then the segments are parallel. If the slopes are different, then the segment, if extended infinitely, will intersect at some point. Let's find the slope of \(\overline{CD}\):
| \(m=\frac{y_2-y_1}{x_2-x_1}\) | Of couse, this is the equation for the slope of a segment. You need to know 2 coordinates of a segment in order to find the slope. Luckily, we already have 2 given |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{0-2}{-4-0}\) | Put the fraction in simplest terms now. |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{-2}{-4}\) | A negative number in the numerator and denominator is the same as dividing two positive numbers. |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{2}{4}\) | 2/4 can be simplified still |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{1}{2}\) | |
Let's find the slope of \(\overline{AB}\):
| \(m=\frac{y_2-y_1}{x_2-x_1}\) | I am reminding you of the slope formula again! |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{AB}=\frac{n-5}{-1-3}\) | We can simplify the denominator, but the numerator will need something else |
| \(m\hspace{1mm}\text{of}\hspace{1mm}\overline{AB}=\frac{n-5}{-4}\) | This is the slope of \(\overline{AB}\). |
What did I say above? I said that parallel lines have the same slope! That means that the slope of \(\overline{CD}\) equals the slope of \(\overline{AB}\):
| \(\frac{-4}{1}*\frac{1}{2}=\frac{n-5}{-4}*\frac{-4}{1}\) | Multiply by -4 on both sides to get rid of all the pesky fractions! |
| \(-2=n-5\) | Add 5 to both sides |
| \(3=n\) | Therefore, the coordinates of point A is (-1,3) |
+2446 Oh, I am sorry. I did not anwer the other questions that you asked. Here we go. Our ultimate goal is to find what value for n makes \(\overline{AB}\hspace{1mm}\perp\hspace{1mm}\overline{CD}\). Just like in the previous problem, find the slope of \(\overline{CD}\). Since I have already found the slope in the previous problem that I answered, I don't need to do that calculation again:
\(m\hspace{1mm}\text{of}\hspace{1mm}\overline{CD}=\frac{1}{2}\)
We already know the slope of \(\overline{AB}\) because we calculated it in the previous problem as well:
\(m\hspace{1mm}\text{of}\hspace{1mm}\overline{AB}=\frac{n-5}{-4}\)
The difference with this problen, however, is that we have to find the value for n that makes the segment perpendicular! For a segment to be perpendicular, its slope must be the opposite reciprocal of the to another segment. In other words, its slope, when multiplied to the other segment must be -1. Of course, we are comparing the slope of \(\overline{AB}\) to the slope of \(\overline{CD}\):
| \(\frac{1}{2}m_{\overline{AB}}=-1\) | Multiply by 2 on both sides to get rid of the pesky fraction |
| \(m_{\overline{AB}}=-2\) | |
Now we know that the slope of \(\overline{AB}=-2\), we can solve for n:
| \(-2=\frac{n-5}{-4}\) | Solve for n. Multiply by -4 on both sides to get rid of the fraction |
| \(8=n-5\) | Add 5 to both sides to isolate n. |
| \(13=n\) | |
Therefore, in order for \(\overline{AB}\perp\overline{CD}\), the coordinate is \(A\hspace{1mm}(-1,13)\)
+2446 I'll also answer your next question about making the segments perpendicular. Find the slope of \(\overline{SR}\) by using the slope formula:
| \(m_{\overline{SR}}=\frac{1-(-1)}{2-(-1)}\) | Subtracting a negative number is the same as adding a positive number |
| \(m_{\overline{SR}}=\frac{1+1}{2+1}\) | Simplify the numerator and denominator. |
| \(m_{\overline{SR}}=\frac{2}{3}\) | The fraction is already in simplest terms |
Let's find the slope of \(\overline{PQ}\):
| \(m_{\overline{PQ}}=\frac{1-(-a)}{a-0}\) | Subtracting a negative is the same as adding a positive |
| \(m_{\overline{PQ}}=\frac{1+a}{a-0}\) | Subtracting a number by 0 is itself. |
| \(m_{\overline{PQ}}=\frac{1+a}{a}\) | |
For \(\overline{PQ}\hspace{1mm}||\hspace{1mm}\overline{SR}\), the slopes must be the same. Therefore, set their slopes equal to each other:
| \(\frac{1+a}{a}=\frac{2}{3}\) | Multiply by a on both sides on the equation to get rid of one of the fractions |
| \(1+a=\frac{2}{3}a\) | Multiply by 3 on both sides to get rid of the other fraction |
| \(3(1+a)=2a\) | Distribute the 3 onto both terms in the parentheses |
| \(3+3a=2a\) | Subtract 3a from both sides |
| \(3=-a\) | Multiply by -1 on both sides to isolate a. |
| \(a=-3\) | |
Therefore, the coordinates are now \(P\hspace{1mm}(-3,1)\hspace{1mm}\text{and}\hspace{1mm}Q\hspace{1mm}(0,3)\).
