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Alona [7]
4 years ago
9

A line passes through (1, 3) and (4, 7). Which line would be perpendicular to this line?

Mathematics
1 answer:
GenaCL600 [577]4 years ago
6 0

We can use the points given to solve for the slope.

Slope formula: y2-y1/x2-x1

7-3/4-1

4/3

Find the y-intercept.

y = mx + b

3 = 4/3(1) + b

3 = 4/3 + b

5/3 = b

The slope of a perpendicular line is the opposite and reciprocal.

4/3 -> -3/4

Write the new equation.

y = -3/4x + 5/3

Best of Luck!

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X + y + w = b
White raven [17]

Answer:

(a) a=6 and b≠\frac{11}{4}

(b)a≠6

(c) a=6 and b=\frac{11}{4}

Step-by-step explanation:

writing equation in agumented matrix form

\begin{bmatrix}1 &1 & 0 &1 &b\\ 2 &3 & 1 &5 &6\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}

now R_{2} =R_{2}-2\times R_{1}

\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 2 & 2&a &1\end{bmatrix}

now R_{4} =R_{4}-2\times R_{2}

\begin{bmatrix}1 &1& 0 &1 &b\\ 0 &1& 1 &3 &6-2b\\ 0& 0 & 1 &1 &4\\ 0& 0 & 0 &a-6 &4b-11\end{bmatrix}

a) now for inconsistent

rank of augamented matrix ≠ rank of matrix

for that  a=6 and b≠\frac{11}{4}

b) for consistent w/ a unique solution

rank of augamented matrix = rank of matrix

  a≠6

c) consistent w/ infinitely-many sol'ns

  rank of augamented matrix = rank of matrix < no. of variable

for that condition

 a=6 and b=[tex]\frac{11}{4}

then rank become 3 which is less than variable which is 4.

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